http://codeforces.com/problemset/problem/870/C

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.

思维题;

问,一个数能拆分成几个(最多)合数?

最小的合数是4,对4取余之后,根据余数判断

//如果余数是0 直接 n/4
//如果余数是1 1可以和两个4 搭配形成一个 9  消耗一个 
//如果余数是2 2可以和一个4 搭配形成一个 6  不消耗 
//如果余数是3 3可以和三个4 搭配形成两个 6,9  消耗一个 
//这个题不能用素数来判断因为 19数素数 但是 可以拆成 4,6,9
//只能特判 

#include <iostream>
#include <cstring>
#include <string>
#include <algorithm>
#include <cstdio>
#include <queue>
#include <stack>
#define ll long long
#define inf 0x3f3f3f3f 
using namespace std; 
//最小的合数是4 对4取余
//如果余数是0 直接 n/4
//如果余数是1 1可以和两个4 搭配形成一个 9  消耗一个 
//如果余数是2 2可以和一个4 搭配形成一个 6  不消耗 
//如果余数是3 3可以和三个4 搭配形成两个 6,9  消耗一个 
//这个题不能用素数来判断因为 19数素数 但是 可以拆成 4,6,9
//只能特判 
int main()
{
	int t;
	ll n;
	cin>>t;
	while(t--)
	{
		cin>>n;
		if(n==1||n==2||n==3||n==5||n==7||n==11)	cout<<"-1"<<endl;
		else if(n%4==0)
		{
			cout<<n/4<<endl;
		 } else if(n%4==1)
		 {
		 	cout<<n/4-1<<endl;
		 }else if(n%4==2)
		 {
		 	cout<<n/4<<endl;
		 }else if(n%4==3)
		 {
		 	cout<<n/4-1<<endl;
		 }
	}	
	return 0;
}
 

 


年轻也需要有所作为( ・´ω`・ )