# GitBook

CodeForces - 606A
 Time Limit: 2000MS Memory Limit: 262144KB 64bit IO Format: %I64d & %I64u

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 colorinto 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 ab 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, xy 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.

```#include<stdio.h>
int main() {
int a,b,c,x,y,z;
while(scanf("%d %d %d",&a,&b,&c)!=EOF) {
scanf("%d %d %d",&x,&y,&z) ;
int sum1=0;
int sum2=0;
if(a>x)
sum1+=(a-x)/2;
else
sum2+=x-a;
if(b>y)
sum1+=(b-y)/2;
else
sum2+=y-b;
if(c>z)
sum1+=(c-z)/2;
else
sum2+=z-c;
if(sum1<sum2)
printf("No\n");
else
printf("Yes\n");
}
return 0;
}```

所以决定恶补英语-.-