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-;
 const double pi=acos(-1.0);
 const int K=1e6+;
 const int mod=1e9+;

 int main(void)
 {
     int n,x,ans;cin>>x;
     while(x--)
     {
         scanf("%d",&n);
         if(n%==)
             ans=n/;
         else if(n%==)
         {
             if(n<)
                 ans=-;
             else if(n==)
                 ans=;
             else
                 ans=(n-)/+;
         }
         else if(n%==)
         {
             if(n<)
                 ans=-;
             else if(n==)
                 ans=;
             else
                 ans=n/;
         }
         else
         {
             if(n<)
                 ans=-;
             else
                 ans=(n-)/+;
         }
         printf("%d\n",ans);
     }
     return ;
 }