Codeforces Round #440 div2 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 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 将一个数分为若干个合数 求出最多可以有多少个  

例 

4 》 4 = 1（本身也是合数）

5 = -1（本身不是合数）

6 》 6 = 1

7 = -1

8 》 4，4 = 2

9 》 9 = 1

19 》 4,6,9 = 3

小于12的数特判

对于大于等于12的数n：

找到的规律是 让n对最小的合数4取余

n%4==0 输出n/4

n%4==1，1与两个4结合成9  也就是输出n/4-1

n%4==2，2与一个4结合成6，也就是输出n/4

n%4==3，三拆为1+2,1与两个结合为9,2与一个4结合为6，也就是输出n/4-1

#include <cstdio>#include <cstring>#include <string>#include <iostream>#include <algorithm>using namespace std;int main(){        int n, m;        scanf("%d", &n);        for (int i = 0; i < n; i ++) {                scanf("%d", &m);                if (m == 1 || m == 2 || m == 3 || m == 5 || m == 7 || m == 11) {                        printf("-1\n");                }                else if (m % 4 == 0) {                        printf("%d\n", m / 4);                }                else if (m % 4 == 1 || m % 4 == 3) {                        printf("%d\n", m / 4 - 1);                }                else if (m % 4 == 2) {                        printf("%d\n", m / 4);                }        }}