if (j & (1<<val)) {
	    // i th of hat, j ranges from 0~2^10
        dp[i][j] = (dp[i-1][j-(1<<val)]+dp[i][j])%(MOD); 
}

const int MOD = 1e9+7;
class Solution {
public:
    int numberWays(vector<vector<int>>& people) {
        vector<vector<int>> hats;
        map<int, vector<int>> mp;
        for (int i = 0; i < people.size(); ++i) {
            for (int j:people[i])
                mp[j].push_back(i);
        }
        for (int i = 1; i <= 40; ++i) {
            if (mp.count(i)) {
                sort(mp[i].begin(), mp[i].end());
                hats.push_back(mp[i]);
            }
        }
        int h = hats.size();
        int p = people.size();
        long long dp[42][(1<<10)+5];
        memset(dp, 0, sizeof(dp));
        dp[0][0] = 1;
        for (int i = 1; i <= h; ++i) {
            for (int j = 0; j < (1<<p); ++j) {
                for (int k = 0; k < hats[i-1].size(); ++k) {
                    int val = hats[i-1][k];
                    if (j & (1<<val)) {
                       		/* j-(1<<val) == j^(1<<val)*/
                            dp[i][j] = (dp[i-1][j^(1<<val)]+dp[i][j])%(MOD);
                    }
                }
                dp[i][j] = (dp[i][j]+dp[i-1][j]) % MOD;
            }
        }
        return dp[h][(1<<p)-1];
    }
};

// Simple solution  time consumption: 432 ms
const int maxn=1e6+5;
char a[]="4",b[]="62";
int s[maxn];
char str[8];
void init(){
    for (int i=1; i<1e6; i++) {
        sprintf(str,"%d",i);
        if (strstr(str, a)!=NULL||strstr(str, b)!=NULL) s[i]=0;
        else s[i]=1;
    }
}
int main(){
    init();
    int m,n;
    while (scanf("%d%d",&m,&n)==2&&(m||n)) {
        int sum=0;
        for (int i=m; i<=n; i++) {
            sum+=s[i];
        }
        printf("%d\n",sum);
    }
    return 0;
}

// digit dp  
//time consumption: 15 ms
int num[10];
int d[10][2];
/* limit indicates whether this interval is like 0~99..9 */
/* cur marks which digit it is now processed */
int dfs(int cur,bool wassix,bool limit){
    int ans=0;
    if(cur==0) return 1;//cur=0 indicates the end of recursion
    /* have we calculated the interval before ? */
    if(!limit&&d[cur][wassix]!=-1) 
        return d[cur][wassix];
    int maxnum=limit==0?9:num[cur];
    /* divide to subproblems */
    for (int i=0; i<=maxnum; i++) {
        /* not-ok */
        if(i==4||(i==2&&wassix)) continue;
        /* move to a lower digit, was six ?,  a limit ?*/
        else ans += dfs(cur-1, i==6, i==maxnum&&limit);
    }
    if(limit==0) d[cur][wassix]=ans;
    return ans;
}
int cal(int n){
    int len=0;
    while (n) {
        num[++len]=n%10;
        n/=10;
    }
    return dfs(len,0,1);
}
int main(){
    int m,n;
    memset(d, -1, sizeof(d));
    while (scanf("%d%d",&m,&n)==2&&(m||n)) {
        printf("%d\n",cal(n)-cal(m-1));//注意是m-1
    }
    return 0;
}

/*
http://acm.hdu.edu.cn/showproblem.php?pid=4352
XHXJ's LIS

LIS + bitmask dp + digit dp
*/
#define ll long long
#define ull unsigned long long

using namespace std;

const int MOD = 1e9+7;
// const int INF = 0x7fffffff;
const int INF = 0x3f3f3f3f;
const int PI = acos(-1.0);

ll L, R, K;
int num[20];
ll dp[20][(1<<10)+5][15];

int decode(int status) {
    int ans = 0;
    while (status) {
        if (status&1) ans++;
        status >>= 1;
    }
    return ans;
}

int encode(int x, int status) { // O(nlogn) LIS solution
    for (int i = x; i < 10; ++i) {
        if (status & (1<<i)) return (status^(1<<i)) | (1<<x); // replace i with x; the pos(i) >= pos(x)
    }
    return status | (1<<x);
}

ll dfs(int pos, int status, int isub, int iszero) {
    if (pos == 0)
        return decode(status)==K; // decode the status to see if it is satisfied
    if (!isub && dp[pos][status][K]!=-1)  // return stored dp value ?
        return dp[pos][status][K];
    int mx = isub==0 ? 9:num[pos];
    ll ans = 0;
    for (int i = 0; i <= mx; ++i) {
        // is the former digits are all zero ? does it have upper bound ? the status transformation ?
        ans += dfs(pos-1, (i==0&&iszero) ? 0:encode(i, status), isub&&i==mx, iszero&&i==0);
    }
    if (!isub) dp[pos][status][K] = ans;
    return ans;
}

ll cal(long long val) {
    int len = 0;
    while (val) {
        num[++len] = val%10;
        val /= 10;
    }
    return dfs(len, 0, 1, 1);
}

int main() {
    int t;
    scanf("%d", &t);
    int kase = 0;
    memset(dp, -1, sizeof(dp));
    while (t--) {
        scanf("%lld%lld%lld", &L, &R, &K);
        printf("Case #%d: %lld\n", ++kase, cal(R)-cal(L-1));
    }
	return 0;
}

/*
POJ 3252
Round Numbers
*/
#define ll long long
#define ull unsigned long long

using namespace std;

const int MOD = 1e9+7;
// const int INF = 0x7fffffff;
const int INF = 0x3f3f3f3f;
const int PI = acos(-1.0);

const int offset = 35;
ll L, R;
int num[35];
ll dp[35][70][2]; // the last dimension is to identify whether the digits before are all zero

ll dfs(int pos, bool waszero, bool isup, int zeros) {
    ll ans = 0;
    // special case for '0' and '1'
    if (pos==1 && waszero && (isup==0||num[pos]==1)) return 0;
    if (pos==0) return (zeros>=0) ? 1:0;
    if (!isup && dp[pos][zeros+offset][waszero]!=-1) return dp[pos][zeros+offset][waszero];
    int mx = isup ? num[pos]:1;
    for (int i = 0; i <= mx; ++i) {
        ans += dfs(pos-1, i==0&&waszero, i==mx&&isup, (i==0&&waszero)==1 ? 0:(zeros-(i==0 ? -1:1)));
    }
    if (!isup) dp[pos][zeros+offset][waszero] = ans;
    return ans;
}

ll cal(long long val) {
    if (val == 0) return 0;
    int len = 0;
    while (val) {
        num[++len] = val%2;
        val >>= 1;
    }
    return dfs(len, 1, 1, 0);
}

int main() {
    memset(dp, -1, sizeof(dp));
    while (scanf("%lld%lld", &L, &R)==2) {
        printf("%lld\n", cal(R)-cal(L-1));
    }
	return 0;
}

ll L, R;
int num[30];
ll dp[20][20][2000];

ll dfs(int pos, int pre, int pivot, bool isup) {
    ll ans = 0;
    if (pre < 0) return 0; // cut this branch!
    if (pos == 0) return pre==0; // is this number balanced ?
    if (dp[pos][pivot][pre]!=-1 && !isup) return dp[pos][pivot][pre];
    int end = isup ? num[pos]:9;
    for (int i = 0; i <= end; ++i) {
        ans += dfs(pos-1, pre+(pos-pivot)*i, pivot, isup && i==end); // calculate weight!
    }
    if (!isup) dp[pos][pivot][pre] = ans;
    return ans;
}

ll cal(long long val) {
    if (val < 0) return 0;
    int len = 0;
    while (val) {
        num[++len] = val%10;
        val /= 10;
    }
    num[0] = 0;
    ll ans = 0;
    for (int i = len; i > 0; --i) { // to select each position as the pivot
        ans += dfs(len, 0, i, 1);
    }
    return ans-(len-1); // 0, 00, 000 this kind of value are calculated more than one times!
}

int main() {
    memset(dp, -1, sizeof(dp));
    int T;
    scanf("%d", &T);
    while (T--) {
        scanf("%lld%lld", &L, &R);
        printf("%lld\n", cal(R)-cal(L-1));
    }
	return 0;
}

#define ll long long
#define ull unsigned long long

using namespace std;

const int MOD = 1e9+7;
// const int INF = 0x7fffffff;
const int INF = 0x3f3f3f3f;
const int PI = acos(-1.0);

ll L, R;
int len;
int num[20];
/* represent the status: which position, is the previous digit one ? 
   is the the digits before contain '13'? the mod of 13 of the previous number
*/
ll dp[15][2][2][15];

ll dfs(int pos, bool wasone, bool isup, bool ok, int m) {
    ll ans = 0;
    if (pos == 0 && ok) {
        return m==0;
    } else if (pos == 0) return 0;
    if (!isup && dp[pos][wasone][ok][m]!=-1) return dp[pos][wasone][ok][m];
    int end = isup ? num[pos]:9;
    // cout << "end: " << end << endl;
    for (int i = 0; i <= end; ++i) {
        ans += dfs(pos-1, i==1, isup&&i==end, (wasone&&i==3)||(ok), (m*10+i)%13);
    }
    if (!isup) dp[pos][wasone][ok][m] = ans;
    return ans;
}

ll cal(long long val) {
    len = 0;
    while (val) {
        num[++len] = val%10;
        val /= 10;
    }
    num[0] = 0;
    ll ans = 0;
    return dfs(len, 0, 1, 0, 0);
}

int main() {
    memset(dp, -1, sizeof(dp));
    while (scanf("%lld", &R) != EOF) {
        printf("%lld\n", cal(R));
    }
	return 0;
}