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>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
using namespace std;

#define debug(x) cout << #x << " is " << x << endl
typedef pair<int, int> pii;
typedef long long ll;
const int INF = 0x3f3f3f3f, N = 2e5 + 15;
const ll MOD = 1e9 + 7;

int t, n, m;
ll a[N];

int main() 
{
	for (int i = 0; i <= 9; ++i) a[i] = 1;
	for (int i = 10; i <= N; ++i) a[i] = (a[i-9] + a[i-10]) % MOD;
	scanf("%d", &t);
	while (t--) {
		scanf("%d%d", &n, &m);
		ll ans = 0;
		while (n) {
			(ans += a[n%10+m]) %= MOD;
			n /= 10;
		}
		printf("%lld\n", ans);
	}
    return 0;
}