You are given an integer n. You have to apply m operations to it.

In a single operation, you must replace every digit d of the number with the decimal representation of integer d + 1. For example, 1912 becomes 21023 after applying the operation once.

You have to find the length of n after applying m operations. Since the answer can be very large, print it modulo 10^9+7.

Input

The first line contains a single integer t (1 ≤ t ≤ 2 ⋅ 10^5) — the number of test cases.

The only line of each test case contains two integers n (1 ≤ n ≤ 10^9) and m (1 ≤ m ≤ 2 ⋅ 10^5) — the initial number and the number of operations. 

Output

For each test case output the length of the resulting number modulo 10^9+7.

Example

Input


5
1912 1
5 6
999 1
88 2
12 100


Output


5
2
6
4
2115

Note

For the first test, 1912 becomes 21023 after 1 operation which is of length 5.

For the second test, 5 becomes 21 after 6 operations which is of length 2.

For the third test, 999 becomes 101010 after 1 operation which is of length 6.

For the fourth test, 88 becomes 1010 after 2 operations which is of length 4.
#include<bits/stdc++.h>
#define ll long long 
using namespace std;
const int mxn=2e5+10,mod=1e9+7;
int dp[12][mxn];
int main()
{
    int arr[]={1,1,1,1,1,1,1,1,1,1,2,2,2,2,2,2,2,2,2};
    for(int i=0;i<=9;i++)
        for(int j=0;j<10;j++)
            dp[i][j]=arr[i+j];
    for(int i=0;i<=9;i++)
        for(int j=10;j<mxn;j++)
            dp[i][j]=(dp[i][j-10]%mod+dp[i][j-9]%mod)%mod;
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int n,m;
        scanf("%d%d",&n,&m);
        int res=0;
        while(n!=0)
        {
            int rem=n%10;
            res=(res%mod+dp[rem][m]%mod)%mod;
            n/=10;
        }
        printf("%d\n",res);
    }
    return 0;
}