Problem: You are given two positive integers n and k. Print the k-th positive integer that is not divisible by n.

For example, if n=3, and k=7, then all numbers that are not divisible by 3 are: 1, 2, 4, 5, 7, 8, 10, 11, 13 .... The 7-th number among them is 10.

Input

The first line contains an integer t (1 ≤ t ≤ 1000) — the number of test cases in the input. Next, t test cases are given, one per line.

Each test case is two positive integers n (2 ≤ n ≤ 10^9) and k (1 ≤ k ≤ 10^9).

Output

For each test case print the k-th positive integer that is not divisible by n.

Example

Input


6
3 7
4 12
2 1000000000
7 97
1000000000 1000000000
2 1


Output


10
15
1999999999
113
1000000001
1

Solution (in PYTHON3):

Solution:
def findNum(n, k):
    if n > k:
        print(str(k))
    elif n == k:
        print(str(k+1))
    elif n < k:
        num = k//n
        i = num*n
        valNeeded = num + k-n*num
        k = valNeeded
        while valNeeded >= n:
            num = k//n
            i += num*n
            valNeeded = num + k-n*num
            k = valNeeded
        if valNeeded > 0:
            i += valNeeded
        print(i)

        # if k%n != 0:
        #     print(k+num+1)
        # elif num%n == 0:
        #     print(k+num + 1)
        # else:
        #     print(k+num + 1)



for _ in range(int(input())):
    n, k = list(map(int, input().split()))
    findNum(n, k)