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;
#define ll long long
#define f(i,a,n) for(ll i=a ;i<n ;i++)
#define fr(i,a,n) for(ll i=n-1;i>=a;i--)
#define pb push_back
#define read(a,n) for(ll i=0 ;i<n ;i++)cin>>a[i]
#define F first
#define S second
#define endl "\n"

bool pali(string s)
{
    for(ll i=0 ;i<s.size()/2 ;i++)
    if(s[i]!=s[s.size()-i-1])
    return false;

    return true;
}

void no(){
	cout<<-1<<endl;
}

void yes(){
	cout<<"YES\n";
}
ll mod=1000000007;
///////////////////////////////////////////////////////////////////

ll dp[10][200005]={0};

ll solve(int n ,int m)
{
	if(m<10-n)
	return 1;
	
	if(dp[n][m]!=0)
	return dp[n][m]%mod;

	else{
		dp[n][m]=((solve(1,m-(10-n))%mod)+(solve(0,m-(10-n)))%mod)%mod;
		return dp[n][m];
	}	
}

int size(int n)
{
	int ans=0;
	while(n>0)
	{
		ans++;
		n/=10;
	}
	return ans;
}

int main()
{
	ios::sync_with_stdio(0);
    cin.tie(0);
    ll t = 1; 
    cin>>t;
    while(t--){
    	int n,m;
    	cin>>n>>m;
    	ll ans=0;
    	while(n>0)
    	{
    		ans+=(solve(n%10,m)%mod);
    		ans%=mod;
    		n/=10;
		}
    	cout<<ans<<endl;
	}
    return 0;
}
