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<iostream>
using namespace std;
#include<vector>
#include<algorithm>
#include<functional>
#include<iterator>
int main()
{
	vector<int> operate;
	for (int i = 0; i < 200001; i++)
	{
		if (i < 9)
			operate.push_back(2);
		if (i == 9)
			operate.push_back(3);
		if (i > 9)
			operate.push_back((operate[i - 9] + operate[i - 10]) % 1000000007);
	}
	int t;
	cin >> t;
	vector<int> answer;
	for (int i = 0; i < t; i++)
	{
		int n, m;
		cin >> n >> m;
		answer.push_back(0);
		while (n)
		{
			if (m - 10 + n % 10 >= 0)
			{
				answer[i] += operate[m - 10 + n % 10];
			}
			else
				answer[i] += 1;
			answer[i] %= 1000000007;
			n /= 10;
		}
		
	}
	copy(answer.begin(), answer.end(), ostream_iterator<int>(cout, "\n"));
	return 0;
}