Alice and Bob


Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 443    Accepted Submission(s): 333

Problem Description


Bob and Alice got separated in the Square, they agreed that if they get separated, they'll meet back at the coordinate point (x, y). Unfortunately they forgot to define the origin of coordinates and the coordinate axis direction. Now, Bob in the lower left corner of the Square, Alice in the upper right corner of the the Square. Bob regards the lower left corner as the origin of coordinates, rightward for positive direction of axis X, upward for positive direction of axis Y. Alice regards the upper right corner as the origin of coordinates, leftward for positive direction of axis X, downward for positive direction of axis Y. Assuming that Square is a rectangular, length and width size is N * M. As shown in the figure:


HDU 5054 Alice and Bob 数学_HDOJ

Bob and Alice with their own definition of the coordinate system respectively, went to the coordinate point (x, y). Can they meet with each other ?

Note: Bob and Alice before reaching its destination, can not see each other because of some factors (such as buildings, time poor).


Input


There are multiple test cases. Please process till EOF. Each test case only contains four integers : N, M and x, y. The Square size is N * M, and meet in coordinate point (x, y). ( 0 < x < N <= 1000 , 0 < y < M <= 1000 ).


Output


If they can meet with each other, please output "YES". Otherwise, please output "NO".


Sample Input


10 10 5 5 10 10 6 6


Sample Output


YES NO 


/*
HDU 5054 数学
两个人的坐标系不同，如果能碰面 必然在矩形的中心位置
2*x==N 2*y==M 
*/
#include<iostream>
#include<stdio.h>
using namespace std;
int main()
{
  int N,M,x,y;
  while(scanf("%d%d%d%d",&N,&M,&x,&y)!=EOF)
  {
    if(2*x==N&&2*y==M)
      printf("YES\n");
    else
      printf("NO\n");
  }
  return 0;
}