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;

const int MAXN = 2e5 + 10;
const int mod = 1e9 + 7;
int dp[MAXN][10];
void solve()
{
	for(int i = 0; i <= 9; i++)
	{
		dp[0][i] = 1;
	}
	for(int i = 1; i < MAXN; i++)
	{
		for(int j = 0; j <= 9; j++)
		{
			if(j == 9)
				dp[i][j] = (dp[i - 1][0] + dp[i - 1][1]) % mod;
			else
				dp[i][j] = dp[i - 1][j + 1];
		}
	} 
}


int main()
{
	ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
	solve();
	int t, m, num;
	cin >> t;
	while(t--)
	{
		num = 0;
		string str;
		cin >> str >> m;
		int z = str.size();
		for(int i = 0; i < z; i++)
		{
			num += dp[m][str[i] - '0'] ;
			num = num % mod;
		}
		cout << num << endl;
	}
	return 0;
}