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:
'''#include <iostream>
#include<string.h>
#include<math.h>
#include<bits/stdc++.h>

using namespace std;





int main() {
	// your code goes here
    int x,y,l,r,i;
    scanf("%d%d%d%d",&x,&y,&l,&r);
    for(i=l;i<=r;i++)
    {
        long long int temp=(x&i)*(y&i);
        printf("%d %lld\n",i,temp);
    }
    
       
	return 0;
}
'''
t = int(input())

for _ in range(t):
	x = input().split()
	n = int(x[0])
	k = int(x[1])
	y = k//(n-1)
	z = n*y
	x = k-((n-1)*y)
	if(x==0):
		x = -1
	ans = z+x
	print(ans)