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 int long long int
#define F first
#define S second 
#define MK make_pair
#define pb push_back
#define mid (l + r) / 2
 
using namespace std ;
 
const int N = 3e5 + 100 , M = 1e9 + 7;
int n , m , k , a[N] , f[10][N] , s[N] ;
 
int32_t main(){
	ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);
	cout << setprecision(30) << fixed ;
	int ttt;
	cin >> ttt;
	f[0][0] = 1 ;
	s[0] = 1 ;
	for(int i = 1 ; i <= 3e5 ; i++){
		f[0][i] = f[9][i - 1] ;
		f[1][i] = f[0][i - 1] + f[9][i - 1];
		f[2][i] = f[1][i - 1] ;
		f[3][i] = f[2][i - 1] ;
		f[4][i] = f[3][i - 1] ;
		f[5][i] = f[4][i - 1] ;
		f[6][i] = f[5][i - 1] ;
		f[7][i] = f[6][i - 1] ;
		f[8][i] = f[7][i - 1] ;
		f[9][i] = f[8][i - 1] ;
		f[1][i] %= M ;
		f[1][i] %= M ;
		f[2][i] %= M ;
		f[3][i] %= M ;
		f[4][i] %= M ;
		f[5][i] %= M ;
		f[6][i] %= M ;
		f[7][i] %= M ;
		f[8][i] %= M ;
		f[9][i] %= M ;
		s[i] = s[i - 1] + f[9][i - 1] ;
		s[i] %= M ;
	}
	while(ttt--){
		cin >> n >> m;
		int ans = 0 ;
		while(n){
			int x = (n % 10) ;
			ans += s[x + m] ;
			n /= 10 ;
		}
		ans %= M ;
		cout << ans << '\n' ;
	}
}
/*
#include <radman/roderc++.h>
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         \\\\\//\       \:::\\\\/      |||||///|[[[*   \\\\\\///      \\\\\//\~~~~~///////'            *@@\/////////////////////"
         \\\\\//\        \___\\/       \\\\\///|*       \\\\\/"       \\\\\//\~~////////'                 **"///////////"""
         \\\\\//\                     |||||///|                       \\\\\//\///////"
         \\\\\//\                     \\\\\///|                       \\\\\//\////"
         \\\\\///                    |||||////|                       \\\\\////"
         \\\\\/"                     \\\\\///*                        \\\\\/"
                                     """""*
*/
