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<bits/stdc++.h>
#include <sys/types.h>
#include <unistd.h>
#define ull unsigned long long int
//#define lli long long int
#define int long long int
#define NUM 1000000007
#define mp make_pair
#define pb push_back
#define ITR ::iterator
#define endl "\n"
#define fast_io ios_base::sync_with_stdio(false);cin.tie(NULL);
using namespace std;
//ull power(ull x, ull n) {
//   if (n == 0)
//       return 1;
//
//   if (n % 2 == 0) {
//       ull y = power(x, n / 2);
//       return (y * y);
//   }
//   else return ((x * pow(x, n - 1)));
//}
int fact(int n)
{
    int ans = 1;
    while (n)
    {
        ans = (ans * n) % NUM;
        n--;
    }
    return ans;
}
const int maxn = 2e5 + 1;
int dp[10][maxn];
 
int rec(int n, int m)
{
    if (m <= 0 or m < (10 - n)) return 1;
 
    if (dp[n][m] != -1) return dp[n][m];
    if (n + m >= 10) return  dp[n][m] = (rec(1, m - 10 + n) + rec(0, m - 10 + n)) % NUM;
    return dp[n][m];
}
signed main()
{
// #ifndef ONLINE_JUDGE
//     freopen("input.txt", "r", stdin);
//     freopen("output.txt", "w", stdout);
// #endif
    fast_io
    int t;
    cin >> t;
    vector<int> v;
    memset(dp, -1, sizeof dp);
    while (t--) {
        v.clear();
        int i, n, m, ans = 0;
        cin >> n >> m;
        while (n)
        {
            ans += rec(n % 10, m);
            n /= 10;
            ans %= NUM;
        }
        cout << ans << endl;
    }
    return 0;
}