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;
typedef long long ll;
typedef double dd;
#define f(i,k,n) for(ll i=k;i<n;i++)
#define all(s) s.begin(),s.end()
#define vec vector<ll>
#define pb push_back
#define IOS ios::sync_with_stdio(0),cin.tie(0),cout.tie(0)
const ll maci=200002;
const ll mod=1e9+7;
ll dp[10][maci];
int main(){
    IOS;
    ll t=1;cin>>t;



    f(i,0,10) dp[i][0]=1;
    f(i,1,maci){
        f(j,0,9){
            dp[j][i]=dp[j+1][i-1];
        }
        dp[9][i]=(dp[0][i-1]%mod+dp[1][i-1]%mod)%mod;
    }
    while(t--){
        ll n,m,ans=0;
        cin>>n>>m;
        while(n>0){
            ll p=n%10;
            n/=10;
            ans+=dp[p][m];
            ans=ans%mod;
        }
        cout<<ans<<endl;
    }
    return 0;
}

