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 <iostream>

const int MOD = 1e9 + 7;
const int MX = 2e5 + 55;
int memo[MX + 1];
int d[10];

int main() {
  std::ios_base::sync_with_stdio(0); std::cin.tie(0);
  int t = 1; scanf("%d", &t);

  d[0] = 1;

  for (int i = 1; i <= MX; i++) {
    int tmp[10] = { 0 };

    for (int j = 0; j <= 9; j++) {
      if (j == 9) {
        tmp[1] = (tmp[1] + d[j]) % MOD;
        tmp[0] = (tmp[0] + d[j]) % MOD;
      } else {
        tmp[j + 1] = d[j];
      }
    }

    for (int j = 0; j <= 9; j++) {
      d[j] = tmp[j];
      memo[i] = (memo[i] + d[j]) % MOD;
    }
  }

  while (t--) {
    int n, m; scanf("%d%d", &n, &m);
    int ans = 0;

    while (n) {
      ans = (ans + memo[m + n % 10]) % MOD;
      n /= 10;
    }

    printf("%d\n", ans);
  }
}