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<iostream>
#include<vector>
#include<queue>
#include<cstring>
#include<cmath>
#include<map>
#include<stack>
#include<cstdio>
#include<algorithm>
using namespace std;
typedef long long ll;
const int inf = 0x3f3f3f3f;
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define fast ios::sync_with_stdio(false), cin.tie(0), cout.tie(0)
const int mod = 1e9 + 7;
ll t, n, m;
int dp[200500];
int main(){
    fast;
    for (int i = 0; i < 9;i++){
        dp[i] = 2;
    }
    dp[9] = 3;
    for (int i = 10; i < 200005;i++){
        dp[i] = (dp[i - 9] + dp[i - 10]) % mod;
    }
    cin >> t;
    while(t--){
        cin >> n >> m;
        ll ans = 0;
        while(n>0){
            int x = n % 10;
            ans += ((m + x < 10) ? 1 : dp[m + x - 10]);
            ans %= mod;
            n /= 10;
        }
        cout << ans << "\n";
    }
    return 0;
}