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.*;

public class Codeforces1513C {
	
	public static int  mod = (int) (1e9 + 7);

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		
		int dp[][] = new int[10][200001];
		for(int i = 0 ; i < 10 ; i++)dp[i] = fill(i,200_001);
		int t=sc.nextInt();
		StringBuilder sb = new StringBuilder();
		for (int tt=0; tt<t; tt++) {
			int n=sc.nextInt(), steps=sc.nextInt();
			long ans=0;
			for (char c:(n+"").toCharArray())
				ans+=dp[c-'0'][steps];
			sb.append(ans%mod + "\n");
		}
		
		System.out.println(sb);
		sc.close();
	}

	private static int[] fill(int num, int j) {
		int count[] = new int[10];
		count[num]++;
		int ans[] = new int[j];
		ans[0] = 1;
		int sum = 1;
		int newcount[] =new int[10];
		for(int p = 1 ; p < j ; p++) {
			for(int i = 0 ; i < 9 ; i++)newcount[i+1] = count[i];
			newcount[0] = count[9];
			newcount[1]=add(newcount[1], count[9]);
			sum=add(sum, count[9]);
			int[] temp=count;
			count=newcount;
			newcount=temp;
			ans[p]=sum;
			
		}
		return ans;
	}
	private static int add(int a, int b) {
		if(a+b >= mod)return a+b-mod;
		return a+b;
	}

}
