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
#define ll long long

const int N = 2e5 + 5, M = 1e9 + 7;
vector<vector<int>> dp;

void pre() {
	dp = vector<vector<int>> (N, vector<int> (10, 1));
	for(int i = 1 ; i < N ; i++) {
		for(int j = 0 ; j < 9 ; j++) {
			dp[i][j] = dp[i - 1][j + 1];
		}
		dp[i][9] = (dp[i - 1][0] + dp[i - 1][1]) % M;
	}
}

void solve() {
	int res = 0;
	int n, m;
	cin >> n >> m;
	while(n) {
		res = (res + dp[m][n % 10]) % M;
		n /= 10;
	}
	cout << res << "\n";
}

signed main() {
	ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
	int t = 1;
	pre();
	cin >> t;
	while(t--)
		solve();
	return 0;
}