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

Copy

1
12

output

Copy

3

input

Copy

2
6
8

output

Copy

1
2

input

Copy

3
1
2
3

output

Copy

-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.

刚开始看错题了，我不知道是英语差，还是cf的英语表达有点奇怪，老是容易看错题，我都怀疑我的四六级是怎么过的，高中英语怎么学的。

刚开始以为是求合数之和等于n的方案数，自闭后才知道是求最多数的合数之和，那么就是一道水题了。

最小合数是4，那么就t=n/4若余数为0，t就是答案。

若余数为1，拿两个四与这个1组成9。

若余数为2，拿一个四组成6，

余数为3，拆成2和1拿一个四组成6，拿两个四组成9

AC代码:

#include<bits/stdc++.h>
using namespace std;
int cal(int n)
{
  int t=n/4;
  int yu=n%4;
  if(yu==0) return t;
  if(yu==1) 
  {
    if(t>=2)  return t-1;
    else return -1;
  }
  if(yu==2) return t;
  if(yu==3) 
  {
    if(t>=3) return t-1;
    else return -1;
  }
}
int main()
{
  int t;
  cin>>t;
  while(t--)
  {
    int n;cin>>n;
    if(n<4) printf("-1\n");
    else  printf("%d\n",cal(n));
  }
}