#include<cstdio>#include<cstring>#include<cstdlib>#include<cctype>#include<cmath>#include<iostream>#include<sstream>#include<iterator>#include<algorithm>#include<string>#include<vector>#include<set>#include<map>#include<stack>#include<deque>#include<queue>#include<list>#define lowbit(x) (x & (-x))const double eps = 1e-8;inline int dcmp(double a, double b){    if(fabs(a - b) < eps) return 0;    return a > b ? 1 : -1;}typedef long long LL;typedef unsigned long long ULL;const int INT_INF = 0x3f3f3f3f;const int INT_M_INF = 0x7f7f7f7f;const LL LL_INF = 0x3f3f3f3f3f3f3f3f;const LL LL_M_INF = 0x7f7f7f7f7f7f7f7f;const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1};const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1};const int MOD = 1e9 + 7;const double pi = acos(-1.0);const int MAXN = 10000 + 10;const int MAXT = 10000 + 10;using namespace std;int dp[10][15];int tmp[10];void init(){    dp[0][0] = 1;    for(int i = 1; i <= 6; ++i){//最多六位数，从低位到高位        for(int j = 0; j < 10; ++j){            for(int k = 0; k < 10; ++k){                if(j != 4 && !(j == 6 && k == 2)) dp[i][j] += dp[i - 1][k];//在dp[i - 1][k]的基础上第i位为j            }        }    }}int solve(int x){//solve函数计算的是1~x-1是否是不幸数的情况，并没有考虑数字x。    int cnt = 0;    while(x){        tmp[++cnt] = x % 10;        x /= 10;    }    tmp[++cnt] = 0;//防越界    int ans = 0;    for(int i = cnt - 1; i >= 1; --i){        for(int j = 0; j < tmp[i]; ++j){//只统计了1~x-1的情况            if(j != 4 && !(tmp[i + 1] == 6 && j == 2)){                ans += dp[i][j];            }        }        if(tmp[i] == 4 || (tmp[i + 1] == 6 && tmp[i] == 2)) break;//一旦当前枚举到的数中含4或62，则后面无论跟什么数字都不符合要求，停止枚举    }    return ans;}int main(){    int n, m;    init();    while(scanf("%d%d", &n, &m) == 2){        if(!n && !m) return 0;        printf("%d\n", solve(m + 1) - solve(n));    }    return 0;}