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.io.*;
import java.util.*;

public class A
{

	FastScanner in;
	PrintWriter out;
	int i = 0, j = 0;


	void solve() {
		/**************START**************/
		int a = in.nextInt(), b = in.nextInt(), c = in.nextInt(), d = in.nextInt(), e = in.nextInt(), f = in.nextInt();
		int wild = 0;
		int diff = 0;
		if (a > d && a > 1)
		{
			diff = a - d;
			if (diff % 2 != 0)
			{
				diff++;
			}
			wild += (diff/2);
			a -= diff;
			
		}
		if (b > e && b > 1)
		{
			diff = b - e;
			if (diff % 2 != 0)
			{
				diff++;
			}
			wild += (diff/2);
			b -= diff;
		}
		if (c > f && c > 1)
		{
			diff = c - f;
			if (diff % 2 != 0)
			{
				diff++;
			}
			wild += (diff/2);
			c -= diff;
		}
		int needed = 0;
		if (d > a)
		{
			needed += (d-a);
		}
		if (e > b)
		{
			needed += (e-b);
		}
		if (f > c)
		{
			needed += (f-c);
		}
		if (wild >= needed)
		{
			out.println("Yes");
		}
		else
		{
			out.println("No");
		}
		











		/***************END***************/
	}

	public static void main(String[] args) {
		new A().runIO();
	}

	void runIO() {
		in = new FastScanner(System.in);
		out = new PrintWriter(System.out);

		solve();

		out.close();
	}

	class FastScanner {
		BufferedReader br;
		StringTokenizer st;

		public FastScanner(File f) {
			try {
				br = new BufferedReader(new FileReader(f));
			} catch (FileNotFoundException e) {
				e.printStackTrace();
			}
		}

		public FastScanner(InputStream f) {
			br = new BufferedReader(new InputStreamReader(f));
		}

		String next() {
			while (st == null || !st.hasMoreTokens()) {
				String s = null;
				try {
					s = br.readLine();
				} catch (IOException e) {
					e.printStackTrace();
				}
				if (s == null)
					return null;
				st = new StringTokenizer(s);
			}
			return st.nextToken();
		}

		//This will empty out the current line (if non-empty) and return the next line down. If the next line is empty, will return the empty string.
		String nextLine() {
				st = null;
				String s = null;
				try {
					s = br.readLine();
				} catch (IOException e) {
					e.printStackTrace();
				}
				return s;
		}

		boolean hasMoreTokens() {
			while (st == null || !st.hasMoreTokens()) {
				String s = null;
				try {
					s = br.readLine();
				} catch (IOException e) {
					e.printStackTrace();
				}
				if (s == null)
					return false;
				st = new StringTokenizer(s);
			}
			return true;
		}

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

		long nextLong() {
			return Long.parseLong(next());
		}

		double nextDouble() {
			return Double.parseDouble(next());
		}
	}
}