Codeforces 560B Gerald is into Art【思维、分类讨论】

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 an
a1 × 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).

Examples

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.



题目大意：

现在有一个大画板两幅画，问你能否将两幅画放到大画板中。

可以对画进行旋转。

思路：

既然可以对画进行旋转，那么我们只要判断所有情况即可。

Ac代码：

#include<stdio.h>
#include<string.h>
#include<iostream>
using namespace std;
int main()
{
int n,m;
while(~scanf("%d%d",&n,&m))
{
int a,b;
int c,d;
scanf("%d%d%d%d",&a,&b,&c,&d);
if((a+c)<=n&&max(b,d)<=m||(a+c)<=m&&max(b,d)<=n)
{
printf("YES\n");
}
else if((a+d)<=n&&max(b,c)<=m||(a+d)<=m&&max(b,c)<=n)
{
printf("YES\n");
}
else if((b+d)<=n&&max(a,c)<=m||(b+d)<=m&&max(a,c)<=n)
{
printf("YES\n");
}
else if((b+c)<=n&&max(a,d)<=m||(b+c)<=m&&max(a,d)<=n)
{
printf("YES\n");
}
else printf("NO\n");
}
}