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;
inline int read(){
	int X=0; bool flag=1; char ch=getchar();
	while(ch<'0'||ch>'9') {if(ch=='-') flag=0; ch=getchar();}
	while(ch>='0'&&ch<='9') {X=(X<<1)+(X<<3)+ch-'0'; ch=getchar();}
	if(flag) return X;
	return ~(X-1);
}
int T,n,m;
typedef long long ll;
ll mod=1e9+7,ans=0ll;
ll f[200050];
int main(){
	cin>>T;
	for(int i=0;i<=8;i++){
			f[i]=2ll;
		}
	for(int i=9;i<=200005;i++){
		if(i<10)f[i]=f[i-9]+1;
		else f[i]=f[i-9]+f[i-10];
		if(f[i]>mod)f[i]%=mod;
	}
	while(T--){
		n=read();m=read();
		ans=0ll;
		while(n){
			int t=n%10;
			if(m>=10-t)ans+=f[m-(10-t)];
			else ans++;
			n/=10;
			if(ans>mod)ans%=mod;
		}
		printf("%d\n",ans%mod);
	}
}