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>
#define FAST ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
#define ll long long int
#define fab(i,a,b) for(i=a;i<=b;i++)
#define fabr(i,a,b) for(i=b;i>=a;i--)
#define f(i,n) for(i=0;i<n;i++)
#define sc second
#define fr first
#define pb push_back
#define mp make_pair
#define mod 1000000007
#define max 1000000000000000000
#define nl "\n"
using namespace std;
ll dp[200001];
void fun()
{
    ll i;
    f(i,200001)
    {
        if(i<=8)
        dp[i]=2;
        else if(i==9)
        dp[i]=3;
        else
        dp[i]=(dp[i-9]+dp[i-10])%mod;
    }
}
int main()
{
    FAST;
    ll t,i,m,z,u,ans,b,c,l,q,p,j,sum,d,n,a,x,y,k,w,h,r;
    t=1;
    char ch;
    string s,aa,bb,dd,rr,tt,ss="";
    fun();
    cin>>t;
    while(t--)
    {
       ans=0;
       cin>>n>>m;
       //cout<<dp[92]<<nl;
       while(n>0)
       {
           k=n%10;
           if(k+m<10)
           ans+=1;
           else
           ans+=dp[k+m-10];
           ans%=mod;
           n=n/10;
       }
       cout<<ans<<nl;
    }  
	return 0;
}
