【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 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:求一个数能分解成为最多多少个合数。根据合数的定义，最小的合数是4 所以肯定是尽量多的分成4啦 那么就会出现不足4的情况 ，分三种情况讨论即可 具体见代码注释

#include <bits/stdc++.h>using namespace std;int solve(int x){    int t = x/4; //最多分成t个4    int rest = x%4; //求出剩余部分    if (rest==0) //刚好整除4 t就是最大值返回        return t;    if (rest==1) //余1 咋办 取出一个4 发现4+1=5 不是合数 那就取出两个 4*2+1=9 是合数 那么就9了    {        if (t>=2)        {            t-=2;        }else            return -1;        t++;        return t;    }    if (rest==2) //同理 余2 取出一个4 发现4+2=6是合数 那就取出一个就行     {        if (t>=1)        {            t--;        }else return -1;        t++;        return t;    }    if (rest==3) //同理 取出3个4 3*4+3=15是合数     {        if (t>=3)        {            t-=3;        }else return -1;        t+=2;        return t; //返回t    }}int main(){    //freopen("F:\\rush.txt","r",stdin);    int q;    scanf("%d",&q);    for (int i = 1;i <= q;i++)    {        int x;        scanf("%d",&x);        printf("%d\n",solve(x));    }    return 0;}