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.
import os,io
input=io.BytesIO(os.read(0,os.fstat(0).st_size)).readline

class Solution:
    def __init__(self):
        #self.cache = {}
        self.mod = 10**9 + 7
        N = 2 * 10**5 + 1
        self.dp = [1] * N
        for i in range(10,N):
            self.dp[i] = (self.dp[i-10] + self.dp[i-9]) % self.mod
        
    
    def getLen(self, x):
        while x >= len(self.dp):
            self.dp.append((self.dp[-10] + self.dp[-9]))
        
        return self.dp[x]
        

    
    def solve(self, n, m):
        ans = 0
        while n > 0:
            dig = n % 10
            n //= 10
            ans = (ans + self.getLen(dig + m)) % self.mod
        
        return ans


        
s = Solution()
numCases = int(input())  
for i in range(1, numCases+1):
    n, m = map(int,input().split())
    ans = s.solve(n, m) 
    print("%d" % ans)