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;
char a[20];
ll dp[200020];
ll cnt1[10];
ll cnt2[10];
ll mod=1e9+7;
int main()
{
	cnt1[0]=1;
	for(int i=1;i<=9;i++){
		cnt1[i]=0;
	}
	dp[0]=1;
	for(int i=1;i<200020;i++){
		if(i%2==1){
			memset(cnt2,0,sizeof(cnt2));
		}else{
			memset(cnt1,0,sizeof(cnt1));
		}
		for(int j=0;j<9;j++){
			if(i%2==1){
				cnt2[j+1]=cnt1[j]%mod;
			}else{
				cnt1[j+1]=cnt2[j]%mod;
			}		
		}
		if(i%2==1){
			cnt2[0]=(cnt2[0]+cnt1[9])%mod;
			cnt2[1]=(cnt2[1]+cnt1[9])%mod;
		}else{
			cnt1[0]=(cnt1[0]+cnt2[9])%mod;
			cnt1[1]=(cnt1[1]+cnt2[9])%mod;
		}
		ll sum=0;
		for(int j=0;j<10;j++){
			if(i%2==1){
				sum=(sum+cnt2[j])%mod;
			}else{
				sum=(sum+cnt1[j])%mod;
			}	
		}
		dp[i]=sum%mod;
	}	
	ll t;
	scanf("%lld",&t);
	while(t--){
		ll n,m;
		memset(a,'\0',sizeof(a));
		scanf("%s%lld",a,&m);
		ll sum=0;
		for(int i=0;i<strlen(a);i++){
			sum=(sum%mod+dp[a[i]-'0'+m]%mod)%mod;
		}
		printf("%lld\n",sum);
	} 
 	return 0;
}

