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.
import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int index = scanner.nextInt();
        while(index-->0){
            int i = scanner.nextInt();
            if(i < 4)
                System.out.println(-1);
            else{
                int fl = getNum(i);
                System.out.println(fl);
            }
           //arr[index] = getNum(i);

        }
       /* for (int i = arr.length-1; i >= 0 ; i--) {
            int i1 = arr[i];
            System.out.println(i1);
        }*/
    }
    public static  int getNum(int sc){
           int itg = sc / 4;//求整数
           int rmd = sc % 4;//求余数
        //合数（除了1和本身还有其它约数 最小的合数 4）  4 6 8 9
        //当余数为 0 答案就是 itg

        /*if(sc < 4){
         return -1;
        }*/

        if(rmd == 0) return itg;


        //当余数为 1 如果 除出来整数大于等于2（除去5以外）那么就必须要两个4 组成9  itg -1
        if(rmd == 1){
            if(itg >= 2) return itg-1;
            else return -1;


        }
        //当余数为 2 必须要拿一个4 组成 6  itg
        if(rmd == 2) return itg;

        //当余数为 3 必须拿一个6组成9 那么这个数itg必须大于等于15
        if(rmd == 3)
        {
           if(itg >= 3) return itg - 1;
           else return -1;

        }
    return 0;
    }
}