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:
//package practice;

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.math.BigInteger;
import java.text.SimpleDateFormat;
import java.time.Instant;
import java.time.LocalDate;
import java.time.LocalDateTime;
import java.time.LocalTime;
import java.util.*;
import java.util.LinkedList;

public class Main {
	

	
	static long countDivisibles(long A, long B, long M)
    {
        
        if (A % M == 0)
            return (B / M) - (A / M) + 1;
 
        
        return (B / M) - (A / M);
    }
 
	
	public static void main(String[] args) throws IOException  {
		Scanner sc=new Scanner(System.in);
		int t=sc.nextInt();
		while(t-->0) {
			long n=sc.nextLong();
			long k=sc.nextLong();
			
			long a=k+1;
			long b=k+(k/n);
			
			long count=0;
			
			while(countDivisibles(a, b,n) !=0) {
				count +=countDivisibles(a, b,n);
				long temp=countDivisibles(a, b,n);
				a=b+1;
				b +=temp;
			}
			
			long res= k + (k/n) + count;
			
			System.out.println(res);
		}
	}
	
}