Technocup 2018 - Elimination Round 2 870C. Maximum splitting

You are given several queries. In the i-th query you are given a single positive integer ni. You are to represent ni as a sum of maximum possible number of composite summands and print this maximum number, or print -1, if there are no such splittings.

An integer greater than 1 is composite, if it is not prime, i.e. if it has positive divisors not equal to 1 and the integer itself.

Input

The first line contains single integer q (1 ≤ q ≤ 105) — the number of queries.

q lines follow. The (i + 1)-th line contains single integer ni (1 ≤ ni ≤ 109) — the i-th query.

Output

For each query print the maximum possible number of summands in a valid splitting to composite summands, or -1, if there are no such splittings.

Examples

input

1
12

output

3

input

2
6
8

output

1
2

input

3
1
2
3

output

-1
-1
-1

Note

12 = 4 + 4 + 4 = 4 + 8 = 6 + 6 = 12, but the first splitting has the maximum possible number of summands.

8 = 4 + 4, 6 can’t be split into several composite summands.

1, 2, 3 are less than any composite number, so they do not have valid splittings.

题意

给出ｎ组询问，问一个数最多由多少复合数组成（复合数是大于一的素数）

题解

最基础的复合数就是４,６,９,然后对m==1||m==2||m==3||m==5||m==7||m==11的时候进行特判。
当m%4等于０的时候说明自然ｍ/4就是最大个数
当m%4等于２的时候那么可以将一个４和余下的２组合成６，那么也是m/4个
当m%4等于１的时候则需要让出两个４和余下的１组成９，所以是m/4+1个
当m%4等于３的时候则需要３个４和余下的３组成一个６一个９，所以也是m/4-1个

CODE

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <cmath>
using namespace std;
int main()
{
    int n,m;
    scanf("%d",&n);
    for(int i=0;i<n;i++)
    {
        scanf("%d",&m);
        if(m==1||m==2||m==3||m==5||m==7||m==11)
            cout << "-1" << endl;
        else
        {
            int x=m%4;
            int y=m/4;
            if(x==2||x==0)
                cout << y << endl;
            else
                cout <<  y-1 << endl;
        }

    }
    return 0;
}