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 N = 300005;
const int mod = 1e9 + 7;
#define ll long long
#define rep(i, a, b) for(int i=(a);i<=(b);++i)
#define per(i, a, b) for(int i=(a);i>=(b);--i)
#define ms(a, b) memset(a, b, sizeof a) 
int n, T, m;
ll f[N + 100];
int main(){
	for(int i=0;i<=9;i++)f[i]=1;
	for(int i=10;i<N;i++)f[i]=(f[i-10]+f[i-9])%mod;
	scanf("%d", &T);
	while(T--){
		scanf("%d%d", &n, &m);
		int ans = 0;
		while(n)ans=(ans+f[m+n%10])%mod, n /= 10;
		printf("%d\n", ans);
	}
}