Problem Description

A balanced number is a non-negative integer that can be balanced if a pivot is placed at some digit. More specifically, imagine each digit as a box with weight indicated by the digit. When a pivot is placed at some digit of the number, the distance from a digit to the pivot is the offset between it and the pivot. Then the torques of left part and right part can be calculated. It is balanced if they are the same. A balanced number must be balanced with the pivot at some of its digits. For example, 4139 is a balanced number with pivot fixed at 3. The torqueses are 4*2 + 1*1 = 9 and 9*1 = 9, for left part and right part, respectively. It's your job
to calculate the number of balanced numbers in a given range [x, y].

 

Input

The input contains multiple test cases. The first line is the total number of cases T (0 < T ≤ 30). For each case, there are two integers separated by a space in a line, x and y. (0 ≤ x ≤ y ≤ 10¹⁸).

 

Output

For each case, print the number of balanced numbers in the range [x, y] in a line.

 

Sample Input

 

Sample Output

#include<cstdio>

#include<cstring>

#include<iostream>

#include<algorithm>

using namespace std;

int bit[];

__int64 dp[][][];

__int64 dfs(int pos,int o,int l,int limit)

{

    //根据题意，要改符合答案的条件，所以当l==0时返回1

    if (pos==-)

        return l==;

    //注意这里，l<0一定要写，因为有一种情况是在pivot的左边<0之后又在左边==0了，这种情况是不符合的，所以这一步不能少

    if (l<)

        return ;

    __int64 &aa=dp[pos][o][l];

    if (!limit&&aa!=-)

        return aa;

    __int64 ans=;

    int end=limit?bit[pos]:;

    for (int i=;i<=end;i++)

    {

        int next=l;

        next+=(pos-o)*i;

        ans+=dfs(pos-,o,next,limit&&i==end);

    }

    if (!limit)

        aa=ans;

    return ans;

}

__int64 sol(__int64 n)

{

    int len=;

    while(n)

    {

        bit[len++]=n%;

        n/=;

    }

    __int64 ans=;

    for (int i=;i<len;i++)

    {

        ans+=dfs(len-,i,,);

    }

    return ans-(len-);

}

int main()

{

    int o;

    cin>>o;

    while(o--)

    {

        memset(dp,-,sizeof(dp));

        __int64 x,y;

        cin>>x>>y;

        cout<<sol(y)-sol(x-)<<endl;

    }

    return ;

}