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
#define str string
#define pb push_back
#define tvvod int t;cin>>t;while(t--)
#define optimizator ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0)
using namespace std;
int main(){
	// freopen (" .txt", "r", stdin);
	// freopen (" .txt", "w", stdout);
	optimizator;
	tvvod{
		int n;
		cin>>n;
		double a=100/n;
		if(a*n==100){
			cout<<a<<endl;
			continue;
		}
		else{
			for(double i=1; i<=100; i++){
				int h=0;
				for(double j=1; j<=i; j++){
					if((j*100)/i==n){
						cout<<i<<endl;
						h=1;
						break;
					}
				}
				if(h==1){
					break;
				}
			}
		}
	}
}