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<stack>
#include<string>
#include<set>
#include<map>
#include<vector>
#include<math.h>
#include<functional>
#include<limits.h>
#include<utility>
#include<queue>
#include<algorithm>
#include<cstring>
#include<iomanip>
#include<bitset>
#include<time.h>
#include<unordered_map>
#define lowbit(x) ((x)&(-x))
typedef long long ll;
typedef long double ld;
using namespace std;
//
const int maxn = 2e5 + 10;
const int maxm = 1e6 + 10;
const int inf = 0x3f3f3f3f;
const ll mod = 1e9+7;
const double eps = 1e-7;
typedef pair<ll, ll> pii;
int dir[4][2] = { {0,1},{0,-1},{-1,0},{1,0} };


ll dp[10][maxn];
ll dfs(int now, int re)
{
	if (re == 0) return 1;
	if (dp[now][re]) return dp[now][re];
	ll ans = 0;
	if (now + re >= 10)
	{
		ans = dfs(1, re - 10 + now) + dfs(0, re - 10 + now);
		ans %= mod;
	}
	else
		ans = dfs(now + re, 0);
	return dp[now][re] = ans;
}
string s;
int m;
int main()
{
	ios::sync_with_stdio(0);
	cin.tie(0), cout.tie(0);
	int T;
	cin >> T;
	while (T--)
	{
		cin >> s >> m;
		ll ans = 0;
		for (int i = 0; i < s.size(); i++)
			ans += dfs(s[i] - '0', m), ans %= mod;
		cout << ans << endl;
	}
	return ~~(0 - 0);
}