Codeforces Round #440 (Div. 2）C. Maximum splitting

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 integern_i. You are to representn_i 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 to1 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 integern_i (1 ≤ n_i ≤ 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.

题目意思 就不多说了吧 

做这道题的时候有人给我讲题意 然后我给理解错了 这道题是让你求分数字最多的分发  然后我还以为是让你求分法的种数

我还以为是整数划分问题

因为要分解的数字最多  所以就变成了用最小的非素数来分解的问题  也就是说 看这个数字中有多少个4 

如果这个数字取余4得1或者3那么需要在答案中减一  因为你需要两个4合并起来在加上这个余数来得到一个非素数 

#include <cstdio>#include <cstring>#include <algorithm>const int inf = 0x7fffffff;const int N = 100005;using namespace std;int array[N];int main(){int _ , n;scanf("%d",&_);while (_ --) {        scanf("%d",&n);        if(n >= 4) {            if(n == 5 || n == 7 || n == 11) {                printf("-1\n");continue;            }            if(n % 4 == 0 || n % 4 == 2) {                printf("%d\n",n/4);            } else if(n % 4 == 1 || n % 4 == 3) {                printf("%d\n",n/4 - 1);            }        }        else {            printf("-1\n");        }}}