CF 55D

Volodya is an odd boy and his taste is strange as well. It seems to him that a positive integer number is beautiful if and only if it is divisible by each of its nonzero digits. We will not argue with this and just count the quantity of beautiful numbers in given ranges.

Input

The first line of the input contains the number of cases t (1 ≤ t ≤ 10). Each of the next t lines contains two natural numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ 9 ·10¹⁸).

Please, do not use %lld specificator to read or write 64-bit integers in C++. It is preffered to use cin (also you may use %I64d).

Output

Output should contain t numbers — answers to the queries, one number per line — quantities of beautiful numbers in given intervals (from l_i to r_i, inclusively).

Examples

Input

1
1 9

Output

9

Input

1
12 15

Output

2

数位dp。
在这道题中离散化是最关键的一步操作，不然dp数组占用内存过大，无法申请。

#include<cstdio>
#include<cstring>

int shu[20];
int lcm[2530];
long long dp[20][50][2530];

void init(){
    memset(dp,-1,sizeof(dp));
    int cnt = 0;
    for(int i=1;i<=2520;i++) // 离散化操作 将所有的因子处理出来
        if(2520%i==0)lcm[i]=++cnt;
    //printf("%d\n",cnt); // 48
}
int gcd(int a, int b){
    return b?gcd(b,a%b):a;
}
long long dfs(int k, int lc, int r, bool shangxian){
    if(k==0)return r%lc==0;
    if(!shangxian && dp[k][lcm[lc]][r]!=-1)
        return dp[k][lcm[lc]][r];
    int maxn = shangxian?shu[k]:9;
    long long cnt = 0;
    for(int i=0;i<=maxn;i++){
        int now=lc;
        if(i)
            now *= i / gcd(i,lc);
        cnt += dfs(k-1,now,(r*10+i)%2520,shangxian&&i==maxn);
    }
    return shangxian?cnt:dp[k][lcm[lc]][r]=cnt;
}
long long solve(long long x){
    int k=0;
    while(x){
        shu[++k] = x%10;
        x/=10;
    }
    return dfs(k,1,0,1);
}
int main(){
    int t;
    long long A, B;
    init();
    scanf("%d",&t);
    while(t--){
        scanf("%I64d %I64d",&A,&B);
        printf("%I64d\n",solve(B)-solve(A-1));
    }
    return 0;
}