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 sys
import math
import itertools
import functools
import collections
import operator
import fileinput
import copy
from collections import *
 
ORDA = 97  # a
def ii(): return int(input())
def mi(): return map(int, input().split())
def li(): return [int(i) for i in input().split()]
def lcm(a, b): return abs(a * b) // math.gcd(a, b)
def revn(n): return str(n)[::-1]
def dd(): return collections.defaultdict(int)
def ddl(): return collections.defaultdict(list)
def sieve(n):
    if n < 2: return list()
    prime = [True for _ in range(n + 1)]
    p = 3
    while p * p <= n:
        if prime[p]:
            for i in range(p * 2, n + 1, p):
                prime[i] = False
        p += 2
    r = [2]
    for p in range(3, n + 1, 2):
        if prime[p]:
            r.append(p)
    return r
def divs(n, start=2):
    r = []
    for i in range(start, int(math.sqrt(n) + 1)):
        if (n % i == 0):
            if (n / i == i):
                r.append(i)
            else:
                r.extend([i, n // i])
    return r
def divn(n, primes):
    divs_number = 1
    for i in primes:
        if n == 1:
            return divs_number
        t = 1
        while n % i == 0:
            t += 1
            n //= i
        divs_number *= t
def prime(n):
    if n == 2: return True
    if n % 2 == 0 or n <= 1: return False
    sqr = int(math.sqrt(n)) + 1
    for d in range(3, sqr, 2):
        if n % d == 0: return False
    return True
def convn(number, base):
    new_number = 0
    while number > 0:
        new_number += number % base
        number //= base
    return new_number
def cdiv(n, k): return n // k + (n % k != 0)
def ispal(s): # Palindrome
    for i in range(len(s) // 2 + 1):
        if s[i] != s[-i - 1]:
            return False
    return True
   
# a = [1,2,3,4,5] -----> print(*a) ----> list print krne ka new way

def function(arr, size,dp):
	if size == 1 or size == 0:
		return 1
	key = ''.join(map(str, arr[:size]))
	if dp[key]:
		return dp[key]
	output = function(arr, size-1, dp)
	key = ''.join(map(str, arr[:size-1]))
	dp[key] = output
	if (arr[size-2]*10 + arr[size-1]) <= 26:
		output += function(arr, size-2, dp)
		key = ''.join(map(str, arr[:size-2]))
		dp[key] = output
	return output

def function(arr,s,n,dp):
	if s == 0:
		return 0
	if n == 0 and s!= 0:
		return s
	if dp[n][s] != 0:
		return dp[n][s]
	if arr[n-1] <= s:
		dp[n][s] = max(arr[n-1], function(arr,s-arr[n-1],n-1,dp), function(arr,s,n-1,dp))
		return dp[n][s]
	else:
		dp[n][s] = function(arr,s,n-1,dp)
		return dp[n][s]
	
	
def main():
	for _ in range(ii()):
		n, k = mi()
		start = 1
		if n > k:
			if n %k == 0:
				end = n
			else:
				end = n + k - n%k
			k = end				
		else:
			end = k
		#print(start, end)
		while start <= end:
			mid = (start+ end)//2
			if mid * n >= k:
				ans = mid
				end = mid -1
			else:
				start = mid +1
		print(ans)
		
main()