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.io.BufferedReader;
import java.io.BufferedWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.Comparator;
import java.util.HashMap;
import java.util.HashSet;
import java.util.LinkedList;
import java.util.PriorityQueue;
import java.util.Queue;
import java.util.StringTokenizer;

public class A
{
	public static void main (String[] args) throws java.lang.Exception
	{
		FastScanner sc = new FastScanner();
			
		int T = sc.nextInt();
outer:	for(int z=0; z<T; z++) {
			
			//Solution begins
			
			int n = sc.nextInt();
			int k = sc.nextInt();
			int x = k / (n-1);
			if((x+k)%n==0) System.out.println(k+x-1);
			else System.out.println(k+x);
			
			
			//Solution ends
		}
	}
	
	public static void reverse(int arr[]) {
		int i=0; 
		int j=arr.length-1;
		while(i<=j) {
			int temp = arr[i];
			arr[i] = arr[j];
			arr[j] = temp;
			i++;
			j--;
		}
	}
	
	public static int gcd(int a, int b) {
		if(b==0) return a;
		return gcd(b,a%b);
	}
	
	static class FastScanner {
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st=new StringTokenizer("");
		String next() {
			while (!st.hasMoreTokens())
				try {
					st=new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			return st.nextToken();
		}
		
		int nextInt() {
			return Integer.parseInt(next());
		}
		
		long nextLong() {
			return Long.parseLong(next());
		}
	}
}