Problem: 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".

Examples

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 2 orange spheres, which is exactly what he needs.

Solution (in JAVA):

Solution:
import java.util.*;
import java.io.*;
import java.math.*;
public class loser
{
    static class InputReader {
        public BufferedReader br;
        public StringTokenizer token;
        public InputReader(InputStream stream)
        {
            br=new BufferedReader(new InputStreamReader(stream),32768);
            token=null;
        }

        public String next()
        {
            while(token==null || !token.hasMoreTokens())
            {
                try
                {
                    token=new StringTokenizer(br.readLine());
                }
                catch(IOException e)
                {
                    throw new RuntimeException(e);
                }
            }
            return token.nextToken();
        }

        public int nextInt()
        {
            return Integer.parseInt(next());
        }

        public long nextLong()
        {
            return Long.parseLong(next());
        }
    }
    static class card{
        int l;
        int r;
        public card(int ch,int i)
        {
            this.l=ch;
            this.r=i;
        }
    }
    static class sort implements Comparator<card>
    {
        public int compare(card o1,card o2)
        {
            if(o1.l!=o2.l)
                return (int)(o1.l-o2.l);
            else
                return (int)(o1.r-o2.r);
        }
    }
    static void shuffle(long a[])
    {
        List<Long> l=new ArrayList<>();
        for(int i=0;i<a.length;i++)
            l.add(a[i]);
        Collections.shuffle(l);
        for(int i=0;i<a.length;i++)
            a[i]=l.get(i);
    }

    /*static long gcd(long a,long b)
    {
    if(b==0)
    return a;
    else
    return gcd(b,a%b);
    }
    static int ans1=Integer.MAX_VALUE,ans2=Integer.MAX_VALUE,ans3=Integer.MAX_VALUE,ans4=Integer.MAX_VALUE;
    static boolean v[]=new boolean[101];
    static void dfs(Integer so,Set<Integer> s[]){
    if(!v[so.intValue()])
    {
    v[so]=true;
    for(Integer h:s[so.intValue()])
    {
    if(!v[h.intValue()])
    dfs(h,s);
    }
    }
    }
    static class Print{
    public PrintWriter out;
    Print(OutputStream o)
    {
    out=new PrintWriter(o);
    }
    }
    static int CeilIndex(int A[], int l, int r, int key) 
    { 
    while (r - l > 1) { 
    int m = l + (r - l) / 2; 
    if (A[m] >= key) 
    r = m; 
    else
    l = m; 
    } 

    return r; 
    } 

    static int LongestIncreasingSubsequenceLength(int A[], int size) 
    { 
    // Add boundary case, when array size is one 
    int[] tailTable = new int[size]; 
    int len; // always points empty slot 
    tailTable[0] = A[0]; 
    len = 1; 
    for (int i = 1; i < size; i++) { 
    if (A[i] < tailTable[0]) 
    // new smallest value 
    tailTable[0] = A[i]; 

    else if (A[i] > tailTable[len - 1]) 
    // A[i] wants to extend largest subsequence 
    tailTable[len++] = A[i]; 

    else
    // A[i] wants to be current end candidate of an existing 
    // subsequence. It will replace ceil value in tailTable 
    tailTable[CeilIndex(tailTable, -1, len - 1, A[i])] = A[i]; 
    } 
    return len; 
    }*/
    /*static int binary(int n)
    {
    int s=1;
    while(n>0)
    {
    s=s<<1;
    n--;
    }
    return s-1;
    }
    static StringBuilder bin(int i,int n)
    {
    StringBuilder s=new StringBuilder();
    while(i>0)
    {
    s.append(i%2);
    i=i/2;
    }
    while(s.length()!=n)
    {
    s.append(0);
    }
    return s.reverse();
    }*/
    public  static void main(String[] args) 
    {
        InputReader sc=new InputReader(System.in);
        int a[]=new int[3];
        int b[]=new int[3];
        for(int i=0;i<3;i++)
            a[i]=sc.nextInt();
        for(int i=0;i<3;i++)
            b[i]=sc.nextInt();
        int weha=0,wehb=0,wehc=0,wen=0;
        if(a[0]>=b[0])
            weha=a[0]-b[0];
            else
            wen+=b[0]-a[0];
        if(a[1]>=b[1])
            wehb=a[1]-b[1];
            else
            wen+=b[1]-a[1];
        if(a[2]>=b[2])
            wehc=a[2]-b[2];
            else
            wen+=b[2]-a[2];
        //System.out.println(weha+" "+wehb+" "+wehc+" "+wen);
        if(((weha/2)+(wehb/2)+(wehc/2))>=wen)
        System.out.println("Yes");
        else
        System.out.println("No");
    }
}