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.
import java.util.Scanner;

public class C {
    static long mod = (long) (1e9 + 7);

    public static void main(String[] args) {
        Scanner scn = new Scanner(System.in);
        long[] dp = new long[(int) 2e5 + 20];
        for (int i = 0; i < 2e5 + 10; i++) {
            if (i < 10) {
                dp[i] = 1;
                dp[i] = dp[i] % mod;
                continue;
            }
            dp[i] = dp[i - 10] + dp[i - 9];
            dp[i] = dp[i] % mod;

        }
        StringBuilder ans = new StringBuilder();
        int cases = scn.nextInt();
        while (cases-- > 0) {
            int n = scn.nextInt();
            int m = scn.nextInt();
            long count = 0;
            while (n > 0) {
                int digit = n % 10;
                count += dp[digit + m];
                count %= mod;
                n /= 10;
            }
            ans.append(count + "\n");
        }

        System.out.println(ans);


    }


}
