CodeForces 610APasha and Stick （拆数字组成长方形）

A. Pasha and Stick

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Pasha has a wooden stick of some positive integer length n. He wants to perform exactly three cuts to get four parts of the stick. Each part must have some positive integer length and the sum of these lengths will obviously be n.

Pasha likes rectangles but hates squares, so he wonders, how many ways are there to split a stick into four parts so that it's possible to form a rectangle using these parts, but is impossible to form a square.

Your task is to help Pasha and count the number of such ways. Two ways to cut the stick are considered distinct if there exists some integer x, such that the number of parts of length x in the first way differ from the number of parts of length x in the second way.

Input

The first line of the input contains a positive integer n (1 ≤ n ≤ 2·109) — the length of Pasha's stick.

Output

The output should contain a single integer — the number of ways to split Pasha's stick into four parts of positive integer length so that it's possible to make a rectangle by connecting the ends of these parts, but is impossible to form a square.

Examples

input

6

output

1

input

20

output

4

Note

There is only one way to divide the stick in the first sample {1, 1, 2, 2}.

Four ways to divide the stick in the second sample are {1, 1, 9, 9}, {2, 2, 8, 8}, {3, 3, 7, 7} and {4, 4, 6, 6}. Note that {5, 5, 5, 5} doesn't work.

题意：

就是给你一个数字，让你拆成四个数字能够构成一个矩形（不能是正方形）。

思路：

如果是偶数数字，注定不可以拆，求出长方形两条边的和，然后除以 2 就可以得到答案，不能够忽略两条边的和为偶数时，就能够组成正方形。

参考代码：

#include<stdio.h>#include<stdlib.h>#include<math.h>#include<string.h>#include<algorithm>#define MYDD 66using namespace std;int main() {int n,ans;while(scanf("%d",&n)!=EOF) {if(n%2==0) {//只有偶数才能够分割n=n/2;//两边长的和ans=0;if(n%2==0)//当边长和为偶数的时候，就存在一个正方形ans=-1;ans+=(n/2);//一个数字 n 的拆分的方式为 n/2 种 printf("%d\n",ans);} elseputs("0");}return 0;}