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>
using namespace std;

template<typename A, typename B> ostream& operator<<(ostream &os, const pair<A, B> &p) { return os << '(' << p.first << ", " << p.second << ')'; } template<typename T_container, typename T = typename enable_if<!is_same<T_container, string>::value, typename T_container::value_type>::type> ostream& operator<<(ostream &os, const T_container &v) { os << '{'; string sep; for (const T &x : v) os << sep << x, sep = ", "; return os << '}'; }
void dbg_out() { cerr << endl; }
template<typename Head, typename... Tail> void dbg_out(Head H, Tail... T) { cerr << "  " << H; dbg_out(T...); }

#ifdef SMIE
#define debug(args...) cerr << "(" << #args << "):", dbg_out(args)
#else
#define debug(args...)
#endif

const int mxm = 2e5 + 20;
const int MOD = 1e9 + 7;

int one[mxm], zero[mxm], dp[mxm];

int main() {
    ios_base::sync_with_stdio(false);   //DON'T mix C and C++ I/O
    //cin.tie(NULL);                      //DON'T use for interactive problem

    dp[0] = 1, zero[0] = 1;
    for (int i = 1; i < mxm; i++) {
        if (i >= 10) {
            one[i] += zero[i - 10];
            one[i] %= MOD;
            zero[i] += zero[i - 10];
            zero[i] %= MOD;
        }
        if (i >= 9) {
            one[i] += one[i - 9];
            one[i] %= MOD;
            zero[i] += one[i - 9];
            zero[i] %= MOD;
        }
        dp[i] = (dp[i - 1] + one[i]) % MOD;
    }

    int tests = 1, testno = 0;
    cin >> tests;
    while (tests--) {
        int n, m;

        cin >> n >> m;

        int ans = 0;
        while (n > 0) {
            ans = (ans + dp[n % 10 + m]) % MOD;
            n /= 10;
        }

        cout << ans << '\n';
    }
}

