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
`112`
output
`3`
input
`268`
output
`12`
input
`3123`
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】：一个数拆分成几个合数之和，求最多能拆分成几个数之和。

【分析】：其实就是除4看余数，显然4要尽量多取，枚举%4余数讨论一下。

【代码】：  ```#include <bits/stdc++.h>
using namespace std;
int main(){
int n,t;
cin>>t;
while(t--){
cin>>n;
if(n<3) puts("-1");
else if(n==5) puts("-1");
else if(n%4==1) cout<<n/4-1<<"\n";
else if(n%4==0) cout<<n/4<<"\n";
else if(n%4==2) cout<<n/4<<"\n";
else if(n==7||n==11) puts("-1");
else cout<<n/4-1<<"\n";
}
return 0;
}```
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