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 int long long
#define INF 1000000000000000000

const int M = 200005;
const int MOD = 1000000007;
int dp[10][M];

int recur(int d,int m)
{
    if(m==0)
    {
        dp[d][m]=1;
        return 1;
    }
    if(dp[d][m]!=-1)
    {
        return dp[d][m];
    }
    if(d==9)
    {
        dp[d][m]=(recur(0,m-1)+recur(1,m-1))%MOD;
    }
    else
    {
        dp[d][m]=recur(d+1,m-1);
    }
    return dp[d][m];
}

int32_t main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    memset(dp,-1,sizeof(dp));
    for(int i=0;i<10;i++)
    {
        for(int j=0;j<M;j++)
        {
            if(dp[i][j]==-1)
                recur(i,j);
        }
    }
    int tt=1;
    cin >> tt;
    while(tt--)
    {
        int n,m;
        cin >> n >> m;
        int ans=0;
        while(n>0)
        {
            int d=n%10;
            ans=(ans+dp[d][m])%MOD;
            n=n/10;
        }
        cout << ans << '\n';
    }
    return 0;
}