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 <stdio.h>
#include <iostream>
#include <vector>
#include <algorithm> // for copy
#include <iterator>  // for ostream_iterator
using namespace std;

const int P = 1000000007;
const int M = 200000;
int answer_for_0[M+10];
// the result for the digit 0 is answer_for_0[m]
// the result for any other digit d is answer_for_0[m+d]

int main (){
    for (int i=0;i<10;i++)
        answer_for_0[i] = 1;
    
    for (int i=10; i<M+10; i++)
        answer_for_0[i] = (answer_for_0[i-10] + answer_for_0[i-9])%P;
    
    int t;
    cin >> t;
    vector<int> answers (t);

    for (int i=0; i<t; i++){
        int n, m;
        cin >>n >>m;
        int ans = 0;
        while(n > 0){
            ans = (ans+ answer_for_0[m+ n%10] )%P ;
            n/=10;
        }
        answers[i] = ans;
    }
    copy ( answers.begin(), answers.end(), ostream_iterator<int>(cout, "\n") );
    return 0;
}
