You are given two integers n and k.

You should create an array of n positive integers a_1, a_2, ..., a_n such that the sum (a_1 + a_2 + ... + a_n) is divisible by k and maximum element in a is minimum possible.

What is the minimum possible maximum element in a?

Input

The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.

The first and only line of each test case contains two integers n and k (1 ≤ n ≤ 10^9; 1 ≤ k ≤ 10^9).

Output

For each test case, print one integer — the minimum possible maximum element in array a such that the sum (a_1 + ... + a_n) is divisible by k. 

Example

Input


4
1 5
4 3
8 8
8 17


Output


5
2
1
3

Note

In the first test case n = 1, so the array consists of one element a_1 and if we make a_1 = 5 it will be divisible by k = 5 and the minimum possible.

In the second test case, we can create array a = [1, 2, 1, 2]. The sum is divisible by k = 3 and the maximum is equal to 2.

In the third test case, we can create array a = [1, 1, 1, 1, 1, 1, 1, 1]. The sum is divisible by k = 8 and the maximum is equal to 1.
import java.io.*;
import java.util.*;
import java.util.Scanner;
public class K
{
    public static void main (String [] args) throws IOException
    {
        Scanner sc=new Scanner (System.in);
        int t;
        t=sc.nextInt();
        sc.nextLine();
        int i;
        int  r[]=new int [t];
        for(i=0;i<t;i++)
        {
            int n,k;
            n=sc.nextInt();
            k=sc.nextInt();
            sc.nextLine();
            if(n==k)
            r[i]=1;
            if(n<k)
            {
                int q=k/n;
                if(k%n==0)
                r[i]=q;
                else
                r[i]=q+1;
            }
            if(n>k)
            {
                 int quotient;
                if(n%k==0)
                r[i]=1;
                
                else
                {
                    quotient=n/k;
                    quotient+=1;
                    k=k*quotient;
                }
                
                
                if(n<k)
                {
                    int u=k/n;
                    if(k%n==0)
                    r[i]=u;
                    else
                    r[i]=u+1;
                }
            
            }
        }
        for(i=0;i<t;i++)
        {
            System.out.println(r[i]);
        }
    }
}
        