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 sys


input = lambda: sys.stdin.readline().rstrip()


def solve_tc(dp):
    mod = 1000000007
    n, m = map(int, input().split())
    cnt = 0
    while n > 0:
        x = int(n % 10)
        if m + x < 10:
            cnt += 1
        else:
            cnt += dp[m + x - 10]
        cnt = int(cnt % mod)
        n //= 10
    return str(cnt)


def dp_maker():
    max_n = 200005
    mod = 1000000007
    dp = {}
    for i in range(10):
        dp[i] = 2
    dp[9] = 3
    for i in range(10, max_n):
        dp[i] = (dp[i - 9] + dp[i - 10]) % mod

    return dp


dp = dp_maker()
t = int(input())
while t > 0:
    t -= 1
    sys.stdout.write(solve_tc(dp))
    sys.stdout.write("\n")
