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 mem(a,b) memset(a,b,sizeof(a))
#define mk make_pair
#define pb push_back
const int inf=0x3f3f3f3f;
const int maxn=2e5+10;
using namespace std;
typedef long long ll;
const ll mod=1e9+7;
void useiostream()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
}
ll num[15]={0};
ll nex[15]={0};
ll dp[15][maxn]={0};
void init1()
{
    ll ans=0;
    for(int i=0;i<=9;i++)
    {
        dp[i][0]=1;
    }
    for(int k=0;k<=9;k++)
    {
        for(int h=0;h<=9;h++)
        {
            if(h==k)num[h]=1;
            else num[h]=0;
        }
        ans=1;
        for(int j=1;j<maxn;j++)
        {
            for(int i=0;i<=9;i++)
            {
                if(i==1)
                {
                    nex[i]=(num[0]+num[9])%mod;
                }
                else if(i==0)nex[0]=num[9];
                else nex[i]=num[i-1];
            }
            ans=(ans+num[9])%mod;
            for(int i=0;i<=9;i++)
            {
                num[i]=nex[i];
            }
            dp[k][j]=ans;
        }
    }
}
int main()
{
    useiostream();
    int T;
    cin>>T;
    init1();
    while(T--)
    {
        string str;
        cin>>str;
        ll ans1=0;
        int m;
        cin>>m;
        for(int i=0;i<str.size();i++)
        {
            ans1=(ans1+dp[str[i]-'0'][m])%mod;
        }
        
        cout<<ans1<<endl;
    }
}