codeforces 560 B. Gerald is into Art （模拟）

B. Gerald is into Art

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Gerald bought two very rare paintings at the Sotheby's auction and he now wants to hang them on the wall. For that he bought a special board to attach it to the wall and place the paintings on the board. The board has shape of ana1 × b1 rectangle, the paintings have shape of a a2 × b2 and a3 × b3 rectangles.

Since the paintings are painted in the style of abstract art, it does not matter exactly how they will be rotated, but still, one side of both the board, and each of the paintings must be parallel to the floor. The paintings can touch each other and the edges of the board, but can not overlap or go beyond the edge of the board. Gerald asks whether it is possible to place the paintings on the board, or is the board he bought not large enough?

Input

The first line contains two space-separated numbers a1 and b1 — the sides of the board. Next two lines contain numbers a2, b2, a3 and b3 — the sides of the paintings. All numbers ai, bi in the input are integers and fit into the range from 1 to 1000.

Output

If the paintings can be placed on the wall, print "YES" (without the quotes), and if they cannot, print "NO" (without the quotes).

Sample test(s)

input

3 2
1 3
2 1

output

YES

input

5 5
3 3
3 3

output

NO

input

4 2
2 3
1 2

output

YES

Note

That's how we can place the pictures in the first test:

And that's how we can do it in the third one.

真搞不懂，为什么这种傻逼题比赛的时候会觉得＂卧槽好难，搞不定啊怎么办怎么办，放弃吧．．＂

那时候真是一点自信也没有啊．．．．．sad

没自信的时候真是连a+b也觉得难

给出三个矩形的长宽，问第一个矩形能否包含后两个．

矩形不可以相交．两个小矩形的边必须要与大矩形平行．

两个小矩形有四种方法，分别是正正，正反，反正，反反

然后两个矩形相对的位置有两种放法，两个左右相邻，或者两个上下相邻．

这样一共有八种放置放法，有一种可以放置便是有解．

/*************************************************************************
    > File Name: code/cf/#313/B.cpp
    > Author: 111qqz
    > Email: rkz2013@126.com 
    > Created Time: Wed 22 Jul 2015 09:52:54 PM CST
 ************************************************************************/

#include<iostream>
#include<iomanip>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<string>
#include<map>
#include<set>
#include<queue>
#include<vector>
#include<stack>
#define y0 abc111qqz
#define y1 hust111qqz
#define yn hez111qqz
#define j1 cute111qqz
#define tm crazy111qqz
#define lr dying111qqz
using namespace std;
#define REP(i, n) for (int i=0;i<int(n);++i)  
typedef long long LL;
typedef unsigned long long ULL;

    int a1,b1,a2,b2,a3,b3;
bool judge (int x2,int y2,int x3,int y3)
{
    if (x2<=a1&&x3<=a1&&y2+y3<=b1)
    return true;
    if (y2<=b1&&y3<=b1&&x2+x3<=a1)
    return true;
    return false;
}
int main()
{
    cin>>a1>>b1>>a2>>b2>>a3>>b3;
    if (judge(a2,b2,a3,b3)||judge(b2,a2,a3,b3)||judge(b2,a2,b3,a3)||judge(a2,b2,b3,a3))
    {
    puts("YES");
    }
    else
    {
    puts("NO");
    }
  
    return 0;
}