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.
//Bismillahir Rahmanir Raheem
#include <bits/stdc++.h>
using namespace std;

typedef long long int ll;
#define fast ios::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL)
#define mod 1000000007
ll dp[200001][11];
ll f(ll m,ll x){
    if(m==0)return 0;
    if(dp[m][x]!=-1)return dp[m][x];
    dp[m][x]=0;
    if(m>=(10-x)){
    	dp[m][x]=(1+(f(m-(10-x),0))%mod+(f(m-(10-x),1))%mod)%mod;
    }
    return dp[m][x];
}



int main()
{
    fast;
    memset(dp,-1,sizeof(dp));
    f(200001,0);
    f(200001,1);
	int t;
	cin>>t;
	while(t--)
	{
	  ll n,m;
      string s;
	  cin>>s>>m;
	  ll ans=0;
	  n=s.size();
	  for(int i=0;i<n;i++){
	  	ll v=s[i]-'0';
	  	ans=(ans+1+f(m,v))%mod;
	  }
	  cout<<ans<<endl;

	}
	return 0;
}