CF-617A Elephant

A. Stones on the Table

time limit per test：1 seconds

memory limit per test：256 megabytes

An elephant decided to visit his friend. It turned out that the elephant's house is located at point 0 and his friend's house is located at point x(x > 0) of the coordinate line. In one step the elephant can move 1, 2, 3, 4 or 5 positions forward. Determine, what is the minimum number of steps he need to make in order to get to his friend's house.

Input

The first line of the input contains an integer x (1 ≤ x ≤ 1 000 000) — The coordinate of the friend's house.

Output

Print the minimum number of steps that elephant needs to make to get from point 0 to point x.

Examples

Input

5

Output

1

Input

12

Output

3

Note

In the first sample the elephant needs to make one step of length 5 to reach the point x.

In the second sample the elephant can get to point x if he moves by 3, 5 and 4. There are other ways to get the optimal answer but the elephant cannot reach x in less than three moves.

import java.util.Scanner;
public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int count=0;
            count += n / 5;
            n %= 5;
            count += n / 4;
            n %= 4;
            count += n / 3;
            n %= 3;
            count += n / 2;
            n %= 2;
            count += n;
            System.out.println(count);
    }
}

判断从0点到目标点的最短步数，不需要考虑每一步走的需要多长，只需要直接从最大的步数开始走，步数count+=n/5，再把距离n%=5得到走完后的n，接下来接着向下运行，若n为100，则第一次计算步数后，n就等于0了，只能在最后的count+=n运行完后输出，因为count+=n这段代码中只可能为0/1