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 int long long

const int max_n = 200005, mod = 1000000007;
int dp[max_n];

signed main(){

    for(int i=0; i<9; i++)dp[i] = 2;
    dp[9] = 3;
    for(int i=10; i<max_n; i++){
        dp[i] = (dp[i-9] + dp[i-10])%mod;
    }

    ios_base::sync_with_stdio(false);
    cin.tie(NULL);

    int t;
    cin>>t;
    while(t--){
        int n, m;
        cin>>n>>m;
        int ans = 0;
        while(n > 0){
            int x = n%10;
            ans += ((m + x < 10) ? 1 : dp[m + x - 10]);
            ans %= mod;
            n/=10;
        }
        cout<<ans<<"\n";
    }
    return 0;
}