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】: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】:

codeforces Round #440 C Maximum splitting【数学/素数与合数/思维/贪心】_#includecodeforces Round #440 C Maximum splitting【数学/素数与合数/思维/贪心】_.net_02
#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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