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 MAX 200000
#define MOD 1000000007
using namespace std;
int dp[11][MAX+10];
int go(int x,int m)
{
	if(dp[x][m]!=-1)	
		return dp[x][m];
	if(m==0)
		return dp[x][m] = 1;
	if(x==9)
		return dp[x][m] = (go(1,m-1)+go(0,m-1))%MOD;
	return dp[x][m] = go(x+1,m-1);
}
int main()
{
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	for(int i=0;i<=9;++i)
		for(int j=0;j<=MAX;++j)
			dp[i][j] = -1;
	int t;
	cin>>t;
	while(t--)
	{
		string s;
		cin>>s;
		int m;
		cin>>m;
		long long ans = 0;
		for(char c:s)
			ans += go(c-'0',m)*1LL, ans %= MOD;
		cout<<ans<<endl;
	}
	return 0;
}