Magic Spheres

Description

Carl is a beginner magician. He has a blue, b violet and c orange magic spheres. In one move he can transform two spheres of the same color into one sphere of any other color. To make a spell that has never been seen before, he needs at least x blue, y violet and z orange spheres. Can he get them (possible, in multiple actions)?

Input

The first line of the input contains three integers a, b and c (0 ≤ a, b, c ≤ 1 000 000) — the number of blue, violet and orange spheres that are in the magician’s disposal.

The second line of the input contains three integers, x, y and z (0 ≤ x, y, z ≤ 1 000 000) — the number of blue, violet and orange spheres that he needs to get.

Output

If the wizard is able to obtain the required numbers of spheres, print “Yes”. Otherwise, print “No”.

Sample Input

Input

4 4 0

2 1 2

Output

Yes

Input

5 6 1

2 7 2

Output

No

Input

3 3 3

2 2 2

Output

Yes

Hint

In the first sample the wizard has 4 blue and 4 violet spheres. In his first action he can turn two blue spheres into one violet one. After that he will have 2 blue and 5 violet spheres. Then he turns 4 violet spheres into 2 orange spheres and he ends up with 2 blue, 1 violet and 2 orange spheres, which is exactly what he needs.

这道题的思路很简单：

就是看前后变化的各个小球的个数。如果大于零，除以二得到的是最多的可以转化的球的个数；如果小于零，证明是需要被转化的个数，加上。如果结果大于等于零，证明上面的小球数量可以转化到下面的小球数量；否则，不能转化。

代码：

#include<stdio.h>
int main()
{
int a,b,c,a1,b1,c1;
while(~scanf("%d%d%d",&a,&b,&c))
{
scanf("%d%d%d",&a1,&b1,&c1);
int sum=0;
a=a-a1;
b=b-b1;
c=c-c1;
if(a>0)
sum+=a/2;
else
sum+=a;
if(b>0)
sum+=b/2;
else
sum+=b;
if(c>0)
sum+=c/2;
else
sum+=c;
if(sum>=0)
printf("Yes\n");
else
printf("No\n");
}
return 0;
}