#include<bits/stdc++.h>
using namespace std;
#define int long long
const int max_n = 200005, mod = 1000000007;
int p[max_n];
signed main() {

    for (int i = 0; i < 9; i ++)p[i] = 2;
    p[9] = 3;
    for (int i = 10; i < max_n; i ++) {
        p[i] = (p[i - 9] + p[i - 10]) % mod;
    }
    ios_base :: sync_with_stdio(false);
    cin.tie(NULL);

    int t;
    cin >> t;
    while (t --) {
        int n, m;
        cin >> n >> m;
        int ans = 0;
        while (n > 0) {
            int x = n % 10;
            ans += ((m + x < 10) ? 1 : p[m + x - 10]);
            ans %= mod;
            n /= 10;
        }
        cout << ans << "\n";  
    }
    return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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.

### Response

```cpp
#include<bits/stdc++.h>
using namespace std;
#define int long long
const int max_n = 200005, mod = 1000000007;
int p[max_n];
signed main() {

    for (int i = 0; i < 9; i ++)p[i] = 2;
    p[9] = 3;
    for (int i = 10; i < max_n; i ++) {
        p[i] = (p[i - 9] + p[i - 10]) % mod;
    }
    ios_base :: sync_with_stdio(false);
    cin.tie(NULL);

    int t;
    cin >> t;
    while (t --) {
        int n, m;
        cin >> n >> m;
        int ans = 0;
        while (n > 0) {
            int x = n % 10;
            ans += ((m + x < 10) ? 1 : p[m + x - 10]);
            ans %= mod;
            n /= 10;
        }
        cout << ans << "\n";  
    }
    return 0;
}

```