B. Petya and Square

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Little Petya loves playing with squares. Mum bought him a square 2n × 2n in size. Petya marked a cell inside the square and now he is solving the following task.

The task is to draw a broken line that would go along the grid lines and that would cut the square into two equal parts. The cutting line should not have any common points with the marked cell and the resulting two parts should be equal up to rotation.

Petya wants to determine whether it is possible to cut the square in the required manner given the sizes of the square side and the coordinates of the marked cell. Help him.

Input

The first line contains three space-separated integers 2n, x and y (2 ≤ 2n ≤ 100, 1 ≤ x, y ≤ 2n), representing the length of a square's side and the coordinates of the marked cell. It is guaranteed that 2n is even.

The coordinates of the marked cell are represented by a pair of numbers x y, where x represents the number of the row and yrepresents the number of the column. The rows and columns are numbered by consecutive integers from 1 to 2n. The rows are numbered from top to bottom and the columns are numbered from the left to the right.

Output

If the square is possible to cut, print "YES", otherwise print "NO" (without the quotes).

Sample test(s)

input

4 1 1

output

YES

input

2 2 2

output

NO

Note

A sample test from the statement and one of the possible ways of cutting the square are shown in the picture:

解题说明：此题是给定一个大正方形中的一个单位块，要求找出一条不与这个单位块有任何公共点的线把该正方形等分为两个形状一样的块。我们都知道如果要二等分一个三角形，无论这条分割线是什么样子的，它肯定会经过正方形的重心，于是这里只要判断这个单位块是不是与重心有公共点即可，也就是判断此块是不是围绕重心的四个小块。

#include <iostream>
#include<algorithm>
#include<cstdio>
#include<cmath>
#include<cstring>
using namespace std;int main()
{int n,x,y;scanf("%d %d %d",&n,&x,&y);if((x==n/2||x==n/2+1)&&(y==n/2||y==n/2+1)){printf("NO\n");}else{printf("YES\n");}return 0;
}

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