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 JAVA):

Solution:
import java.util.Scanner;
public class MyClass  {
	public static void main(String[] args)  {
		int t;
		long[] n,k;
		long count = 0;
		long ans, div = 0, newDiv = 0;
		Scanner sc = new Scanner(System.in);
		t = sc.nextInt();
		n = new long[t];
		k = new long[t];
		for(int i =  0; i < t;i++){
			n[i] = sc.nextInt();
			k[i] = sc.nextInt();
			// System.out.println(n);
			// System.out.println(k);
		}
		for(int i = 0; i < t; i++){
			div = 0;
			newDiv = 0;
			ans = k[i] + newDiv - div;
			do{
				ans = ans + newDiv - div;
				div = newDiv;
				if (newDiv == 0){
					div = ans / n[i];
					ans += div;
				}
				/*
				else{
					ans = ans + newDiv - div;
					div = ans / n[i];
				}
				*/
				newDiv = ans / n[i];
				// System.out.println(ans);
			}while(newDiv != div);
			/*
			long j = 0;
			while(count < k[i]){
				j++;
				// System.out.println(j);
				if(j%n[i] != 0){
					count ++;
					// System.out.println(count);
				}
			}
			count = 0;
			*/
			System.out.println(ans);
		}
	}
}