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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.*;

public class Practice {

	public static void main(String[] args) throws Exception {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		PrintWriter pw=new PrintWriter(System.out);
		
		long[] arr = new long[200025];
		for (int i = 0; i < 10; i++)
			arr[i] = 1;
		
		int t = Integer.parseInt(br.readLine());
		
		long mod = (long) Math.pow(10, 9) + 7;
		
		for (int i = 10; i < 200025; i++)
			arr[i] = (arr[i - 10] + arr[i - 9]) % mod;
		
		while (t-- > 0) {
			String[] s1 = br.readLine().split(" ");
			long n = Long.valueOf(s1[0]);
			int k = Integer.valueOf(s1[1]);
			long ans = 0;
			while (n > 0) {
				ans = (ans+arr[(k + (int)(n % 10))])%mod;
				n = n / 10;
			}
		//	System.out.println("888888888888");
			pw.println(ans%mod);
		}
		pw.close();
	}
}
