Codeforces Round #440 (Div. 2, based on Technocup 2018 Elimination Round 2) C. Maximum splitting

地址：

题目：

C. Maximum splitting

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

思路：

　　最小合数是4，所以用4去凑就行了

#include <bits/stdc++.h>

using namespace std;

#define MP make_pair
#define PB push_back
typedef long long LL;
typedef pair<int,int> PII;
const double eps=1e-8;
const double pi=acos(-1.0);
const int K=1e6+7;
const int mod=1e9+7;


int main(void)
{
    int n,x,ans;cin>>x;
    while(x--)
    {
        scanf("%d",&n);
        if(n%4==0)
            ans=n/4;
        else if(n%4==1)
        {
            if(n<9)
                ans=-1;
            else if(n==9)
                ans=1;
            else
                ans=(n-13)/4+2;
        }
        else if(n%4==2)
        {
            if(n<6)
                ans=-1;
            else if(n==6)
                ans=1;
            else
                ans=n/4;
        }
        else
        {
            if(n<15)
                ans=-1;
            else
                ans=(n-15)/4+2;
        }
        printf("%d
",ans);
    }
    return 0;
}