Magic Spheres

D - Magic Spheres

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

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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 02 1 2

Output

Yes

Input

5 6 12 7 2

Output

No

Input

3 3 32 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.

有趣的思维题。

可以这样想:对于一种颜色，若它的初始个数大于最终它需要具备的个数，则多出来的那几个(设为n)就可以用来转化为别的颜色，注意，多出来的若为奇数，则就会多出一个无法利用，若刚好为偶数，则可以转化n/2个其它的颜色。若它的初始个数小于最终它需要具备的个数，则这种颜色需要被转化，并且少几个(设为m)，就需要m个别的颜色转化来的。

因此，条件判断为如果(n/2-m)>=0，则可以转化成功，若(n/2-m)<0,则失败。

#include<stdio.h>int main(){    long long a,b,c,x,y,z,sum=0;    scanf("%lld%lld%lld%lld%lld%lld",&a,&b,&c,&x,&y,&z);    if((a-x)>0)        sum+=(a-x)/2;    else        sum+=a-x;    if((b-y)>0)        sum+=(b-y)/2;    else        sum+=b-y;    if((c-z)>0)        sum+=(c-z)/2;    else        sum+=c-z;    if(sum>=0)        printf("Yes\n");    else        printf("No\n");    return 0;}