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;

typedef long long ll;

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(NULL);
    int big = pow(10, 9) + 7, size = 200000;
    int tt;
    cin >> tt;
    vector<ll> dp(size+1);
    dp[0] = 1;
    for (int i = 0; i < size; i++)
    {
        if (i + 9 < size)
            dp[i + 9] = (dp[i + 9] + dp[i]) % big;
        if (i + 10 < size)
            dp[i + 10] = (dp[i + 10] + dp[i]) % big;
    }
    for (int i = 1; i < size; i++)
    {
        dp[i] = (dp[i] + dp[i - 1]) % big;
    }
    while (tt--)
    {
        string s;
        ll m, ans = 0;
        cin >> s >> m;
        ans = s.size();
        vector<ll> num(10);
        for (int i = 0; i < s.size(); i++)
        {
            int n = s[i] - '0';
            num[n]++;
        }
        for (int i = 0; i < 10; i++)
        {
            if (m < (10 - i))
                continue;
            ans = (ans + num[i] * dp[m - (10 - i)]) % big;
        }
        cout << ans << "\n";
    }

    return 0;
}