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 mp make_pair
#define s second
#define pb push_back
#define eb emplace_back
#define all(v) v.begin(),v.end()
#define mx(a,b) a>b?a:b
#define mn(a,b) a<b?a:b
using namespace std;
const int maxn = 5+2e5;
typedef long long ll;
const int mod = 1e9+7; 
int dp[maxn];
int main(){
	ios::sync_with_stdio(0);
	cin.tie(0);
	int t;
	cin >> t;
	for(int i = 0;i < 9;i++)dp[i] = 2;
	dp[9] = 3;
	for(int i = 10;i < maxn;i++){
		dp[i] = (dp[i-9] + dp[i-10]) % mod;
	}
	while(t--){
		ll sol = 0;
		string k;
		int m;
		cin >> k >> m;
		for(auto i: k){
			if(m + (int)(i-'0') < 10)sol++;
			else sol += dp[m-10+(int)(i-'0')];
			sol %= mod;
		}
		cout << sol << "\n";

	}
}