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.*;
import java.io.*;

public class problem714_c {

	static StreamTokenizer in;

	static int nextInt() throws Exception {
		in.nextToken();
		return (int) in.nval;
	}

	static String next() throws Exception {
		in.nextToken();
		return (String) in.sval;
	}
	static long[][]dp=new long[10][200_000+1];
	static int mod=1_000_000_000+7;
	public static void main(String[] args) throws Exception {
		//in = new StreamTokenizer(new BufferedReader(new FileReader("test.in")));
		    in=new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));
		for(int i=0;i<10;i++) fill(i);
		int t=nextInt();
		StringBuilder ans=new StringBuilder();
		for(int i=0;i<t;i++) ans.append(solve()).append("\n");
		System.out.println(ans);
	}
	static String solve() throws Exception{
		int n=nextInt(), m=nextInt();
		long ans=0;
		while(n>0) {
			ans=(ans+dp[n%10][m])%mod;
			n/=10;
		}
		return ""+ans;
	}
	static void fill(int num) {
		int m=200_000;
		long []curr=new long[10];
		curr[num]++;
		dp[num][0]=1;
		for(int i=1;i<=m;i++) {
			long []temp=new long[10];
			for(int j=1;j<=9;j++) {
				temp[j]=curr[j-1];
			}
			temp[0]=(temp[0]+curr[9])%mod; temp[1]=(temp[1]+curr[9])%mod;
			curr=temp;
			for(long val:curr) dp[num][i]=(dp[num][i]+val)%mod;
		}
	}
}

