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>
using namespace std;
typedef long long ll;
typedef vector<ll> vi;
typedef deque<ll> di;
typedef pair<ll,ll> pi;
#define pb push_back
#define mp make_pair
#define mh make_heap
#define pf push_front
#define IOS ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define INF 1000000000
#define FOR(i,a,b) for(ll i=a;i<b;i++)
const ll mod=1e9+7;

int main()
{
    //freopen("contest.txt","r",stdin);
    //freopen("contest.txt","w",stdout);
    IOS
    ll t;
    cin>>t;
    while(t--)
    {
        ll k;
        cin>>k;
        ll r=100-k; ll total=k+r;
        ll s1=k,s2=r;
        for(ll i=2; i<total ; i++)
        {
            if(k%i==0 && r%i==0)
            {
                if(k/i < s1)
                    s1=k/i;
                if(r/i < s2)
                    s2= r/i;
            }
        }
        if(s1==0 || s2==0)
            cout<<1<<endl;
        else
            cout<<s1+s2<<endl;

    }

    return 0;
}
