Gym_100971G_Repair

Alex is repairing his country house. He has a rectangular metal sheet of size a × b. He has to cut two rectangular sheets of sizes a₁ × b₁ and a₂ × b₂ from it. All cuts must be parallel to the sides of the initial sheet. Determine if he can do it.

Input

The first line contains two space-separated integers a and b (1 ≤ a, b ≤ 10⁹) — the sizes of the initial sheet.

The second line contains two space-separated integers a₁ and b₁ (1 ≤ a₁, b₁ ≤ 10⁹) — the sizes of the first sheet to cut out.

The third line contains two space-separated integers a₂ and b₂ (1 ≤ a₂, b₂ ≤ 10⁹) — the sizes of the second sheet to cut out.

Output

Output «YES» (without quotes), if it's possible to cut two described sheets from the initial sheet, or «NO» (without quotes), if it's not possible.

Example

Input

12 2014 75 6

Output

YES

Input

12 2011 920 7

Output

NO

题意：

给你一个a*b的矩形，问你能否在其中割取a1*b1和a2*b2的两个矩形

解：

一共有8中情况，每种情况考虑到了就能写了。一开始思路不太清晰，导致WA了三次。

代码：

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;bool judge(int x1,int y1,int x2,int y2,int a,int b){    if(x1<=a&&y1<=b)    {        if(x1+x2<=a&&max(y1,y2)<=b)        return true;        if(y1+y2<=b&&max(x1,x2)<=a)        return true;        if(x1+y2<=a&&max(y1,x2)<=b)        return true;        if(y1+x2<=b&&max(x1,y2)<=a)        return true;    }    return false;}                                 //默认a为长边，b为短边，就有8种情况，画一下图就都懂了。int main(){    int a,b;    int a1,b1,a2,b2;    while(~scanf("%d%d",&a,&b))    {        int temp=min(a,b);        a=max(a,b);        b=temp;        scanf("%d%d",&a1,&b1);        scanf("%d%d",&a2,&b2);        temp=min(a1,b1);        a1=max(a1,b1);        b1=temp;        temp=min(a2,b2);        a2=max(a2,b2);        b2=temp;        if(judge(a1,b1,a2,b2,a,b)||judge(b1,a1,a2,b2,a,b))        printf("YES\n");        else        printf("NO\n");    }    return 0;}