Codeforces Round #335 (Div. 2) A

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 test(s)

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

Note

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 2orange spheres, which is exactly what he needs.

题意:有 a b c 三种东西，2个a可以换一个b或一个c，其他种类也是相同的换法问可不可以换成想要的种类数量

如果想要的比以前的少了，绝对是两个拿去换一个了，所以是（a-x）/2看能换多少个；如果多了，绝对是其他种类两个换它一个了，那我就减去能换的

#include<stdio.h>
//#include<bits/stdc++.h>
#include<string.h>
#include<iostream>
#include<math.h>
#include<sstream>
#include<set>
#include<queue>
#include<map>
#include<vector>
#include<algorithm>
#include<limits.h>
#define inf 0x3fffffff
#define INF 0x3f3f3f3f
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define LL long long
#define ULL unsigned long long
using namespace std;
int i,j;
int n,m;
int sum,ans,flag;
//int a[1000000];
int t;
int a,b,c;
int x,y,z;
int main()
{
    cin>>a>>b>>c;
    cin>>x>>y>>z;
    sum=0;
    if(a-x>=0)
    {
        sum+=(a-x)/2;
    }
    else if(a-x<0)
    {
        sum-=(x-a);
    }
    if(b-y>=0)
    {
        sum+=(b-y)/2;
    }
    else if(b-y<0)
    {
        sum-=(y-b);
    }
    if(c-z>=0)
    {
        sum+=(c-z)/2;
    }
    else if(c-z<0)
    {
        sum-=(z-c);
    }
    if(sum>=0)
    {
        puts("Yes");
    }
    else
    {
        puts("No");
    }
    return 0;
}