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>
using namespace std;

#define MOD 1000000007
#define MAXM 200015

long long y;

int pd[MAXM];
int solve(int x){
  if(x < 0) return 0;
  if(pd[x] == -1){
    if(x <= 9) pd[x] = 1;
    else{
      y = (solve(x - 10) + solve(x - 9));
      if(y > MOD) y -= MOD;
      pd[x] = y;
    }
  }
  return pd[x];
}

int main(){
  int t; scanf("%d", &t);
  memset(pd, -1, sizeof pd);
  solve(MAXM - 1);
  while(t--){
    char n[10]; scanf("%s", n);
    int m; scanf("%d", &m);

    long long ans = 0;
    for(int i = 0; n[i]; ++i){
      int z = n[i] - '0';
      ans += solve(m + z);
      ans %= MOD;
    }
    printf("%lld\n", ans);
  }
}
