Time Limit: 2 Seconds      Memory Limit: 65536 KB


Give you two circles and a rectangle, your task is to judge wheather the two circles can be put into the rectangle with no part of circles outside the retangle.

Input

There are multiple test cases. In every test cast, there are four float-point numbers:

a, b, r1, r2

where, a and b are two sides of the rectangle, r1 and r2 are radii of the two circls.

Output

Print a "Yes", if the circles can be put into the rectangle. Otherwise, print a "No".

You can safely assume x < y, where x and y are float-point numbers, if x < y + 0.01.

Sample Input

5 4 1 1
5 4 1.5 2

Sample Output

Yes
No

做点水题,舒缓心情,享受人生

简单计算几何

大概没什么好解释的,放进去两个圆以后,根据它们的“横纵方向最大间隔距离”和“半径之和”来判断能否装下。

 /*by SilverN*/
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
using namespace std;
const int mxn=;
double a,b,r1,r2;
int main(){
while(scanf("%lf%lf%lf%lf",&a,&b,&r1,&r2)!=EOF){
if(a<b)swap(a,b);
if(r1<r2)swap(r1,r2);
if(b<=r1*){printf("No\n");continue;}
double x1=a-r1-r2;
double y1=b-r1-r2;
double r=r1+r2;
if(x1*x1+y1*y1-r*r>=0.01)
printf("Yes\n");
else printf("No\n");
}
return ;
}

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