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>

#define ll long long
#define for0() for(int i = 0; i < n; i++)
#define for1() for(int i = 1; i < n; i++)
#define forv(c, k) for(int i = c; i < k; i++)
#define vi vector<int>
#define endl '\n'
#define mod 1000000007

using namespace std;

int dp[12][200010];



int main()
{
	cin.tie(NULL);
	ios_base::sync_with_stdio(false);
	
	for(int i = 0; i < 10; i++)
	{
		dp[i][0] = 1;
	}
	
	for(int i = 1; i < 200002; i++)
	{
		for(int j = 0; j < 10; j++)
		{
			if(j == 9)
			{
				dp[j][i] = (dp[1][i-1] + dp[0][i-1])%mod;
				break;
			}
			dp[j][i] = dp[j+1][i-1];

		}
	}

	
	ll t, n, m, p, q, temp; cin >> t;
	
	
	while(t--)
	{
		cin >> n >> m;
		ll ans = 0;

		

		int nums;

		while(n)
		{
			nums = n%10; n/=10;

			ans = (ans + dp[nums][m])%mod;
		}

		cout << ans << endl;
	}


}

