Beautiful numbers

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).

Example

Input

1
1 9

Output

9

Input

1
12 15

Output

2

DP好难2333

#include<iostream>
#include<stdio.h>
#include<algorithm>
#include<string.h>
using namespace std;


const int mod=2520;
typedef long long LL;

long long dp[25][mod][50];

int index[2530],bit[25];

void init()
{
    int num=0;
    for(int i=1;i<=mod;i++)
    {
        if(mod%i==0)
            index[i]=num++;
    }
}

int gcd(int a,int b)
{
    if(b==0)return a;
    else return gcd(b,a%b);
}

int lcm(int a,int b)
{
    return a/gcd(a,b)*b;
}

LL dfs(int pos,int presum,int prelcm,bool flag)
{
    if(pos==-1)return presum%prelcm==0;
    if(!flag&&dp[pos][presum][index[prelcm]]!=-1)
        return dp[pos][presum][index[prelcm]];
    LL ans=0;
    int up=flag?bit[pos]:9;
    for(int i=0;i<=up;i++)
    {
        int nowsum=(presum*10+i)%mod;
        int nowlcm=prelcm;
        if(i)nowlcm=lcm(nowlcm,i);
        ans+=dfs(pos-1,nowsum,nowlcm,flag&&i==up);
    }
    if(!flag)dp[pos][presum][index[prelcm]]=ans;
    return ans;
}

LL solve(LL x)
{
    int len=0;
    while(x)
    {
        bit[len++]=x%10;
        x/=10;
    }
    return dfs(len-1,0,1,true);
}

int main()
{
    int T;
    init();
    memset(dp,-1,sizeof(dp));
    scanf("%d",&T);
    while(T--)
    {
        LL r,l;
        scanf("%I64d%I64d",&l,&r);
        printf("%I64d\n",solve(r)-solve(l-1));
    }
    return 0;
}