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.
#include<bits/stdc++.h>
#define ll long long
using namespace std;

bool check_prime(int n){
	if(n<=1){
		return false;
	}
	int flag=1;
	for(int i=2;i<=sqrt(n);i++){
		if(n%i==0){
			flag=0;
			break;
		}
	}
	return flag==1;
}

int gcd(int a,int b){
	if(a==0){
		return b;
	}
	return gcd(b%a,a);
}

int main() {
	int t;
	cin>>t;
	while(t--){
	 int k;
	 cin>>k;
	 cout<<100/gcd(k,100)<<endl;
	}
	return 0;
}