You have an initially empty cauldron, and you want to brew a potion in it. The potion consists of two ingredients: magic essence and water. The potion you want to brew should contain exactly k\ \% magic essence and (100 - k)\ \% water.

In one step, you can pour either one liter of magic essence or one liter of water into the cauldron. What is the minimum number of steps to brew a potion? You don't care about the total volume of the potion, only about the ratio between magic essence and water in it.

A small reminder: if you pour e liters of essence and w liters of water (e + w > 0) into the cauldron, then it contains (e)/(e + w) ⋅ 100\ \% (without rounding) magic essence and (w)/(e + w) ⋅ 100\ \% water.

Input

The first line contains the single t (1 ≤ t ≤ 100) — the number of test cases.

The first and only line of each test case contains a single integer k (1 ≤ k ≤ 100) — the percentage of essence in a good potion.

Output

For each test case, print the minimum number of steps to brew a good potion. It can be proved that it's always possible to achieve it in a finite number of steps.

Example

Input


3
3
100
25


Output


100
1
4

Note

In the first test case, you should pour 3 liters of magic essence and 97 liters of water into the cauldron to get a potion with 3\ \% of magic essence.

In the second test case, you can pour only 1 liter of essence to get a potion with 100\ \% of magic essence.

In the third test case, you can pour 1 liter of magic essence and 3 liters of water.
'''input
3
3
100
25



'''

from bisect import bisect_left as bl
from random import randint as R
RI = lambda : [int(_x) for _x in raw_input().split()]


# per = 0.0
# profit = 0.0
# avg_p = 0
# loss = 0.0
# avg_l = 0
# while True:
# 	try:
# 		s = raw_input().strip().split()
# 		if s[0]=="Time":
# 			cp = (float(s[3]) - float(s[2]))/float(s[2])
# 		else:
# 			cp = (float(s[2]) - float(s[1]))/float(s[1])
# 		cp*=100
# 		per+=cp
# 		if cp>0:
# 			profit += cp
# 			avg_p += 1
# 		else:
# 			loss += cp
# 			avg_l += 1
# 	except:
# 		break

# print profit,loss,per+(avg_p+avg_l)*-0.0





def gcd(a,b):
	if b==0:
		return a
	return gcd(b,a%b)

for _i in range(input()):
	n = input()
	gc = gcd(n,100)
	print 100/gc

