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;
typedef long long ll;

const int N = 233333;
const ll mod = 1e9 + 7;

ll dp[10][N];
int main() {
    int m = 200011;
    for(int i = 0; i <= 9; ++i) dp[i][0] = 1;
    for(int i = 1; i <= m; ++i) {
        for(int d = 0; d <= 9; ++d) {
            if(d <= 8) {
                dp[d][i] = dp[d+1][i-1];
            } else {
                dp[d][i] = (dp[1][i-1] + dp[0][i-1]) % mod;
            }
        }
    }
    int T, n, k;
    cin >> T;
    while(T--) {
        scanf("%d %d", &n, &k);
        int x = n;
        ll ans = 0;
        while(x) {
            int bit = x % 10;
            x /= 10;
            ans = (ans + dp[bit][k]) % mod;
        }
        printf("%lld\n", (ans+mod) % mod);
    }
}