870C - Maximum splitting

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.

题意：给你一个数n，要你求最多能够把该数拆分成多少个合数。

题解：通过观察我们发现，当一个数按要求可分的时候，它最多只能分成4,6,9，因此我们就可以有想法：6=4+2,9=4+4+1。我们只需对给的n求余即可，过程看代码

代码：

#include <iostream>
using namespace std;
int main(){
	int T;
	cin>>T;
	while(T--){
		int n;
		cin>>n;
		int mod=n%4,a=n/4,ans;
		if(mod==0) ans=a;
		else if(mod==1){
			if(a>1) ans=a-1;
			else ans=-1;
		}
		else if(mod==2){
			if(a>0) ans=a;
			else ans=-1;
		}
		else if(mod==3){
			if(a>2) ans=a-1;
			else ans=-1;
		}
		cout<<ans<<endl;
	}
	return 0;
}