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:
import math
t = int(input())
for t in range(0, t):
    n, k = [int(x) for x in input().split()]
    # Think of it as grouping numbers in groups of 'n'. In each group you have 'n-1' numbers not divisible by n.
    # So if we need to find the kth number not divisible by 'n' we need to skip the appropriate number of groups.
    # And then find the position 'k' in the next group
    # Keep in mind that it is possible that k is divisible by k-1 in which case we would have to shift the pointer
    # one position to the left
    if k % (n-1) == 0:
        x = (n * math.floor(k / (n-1))) - 1
    else:
        x = (n * math.floor(k / (n-1))) + (k % (n-1))
    print(x)