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 CPP):

Solution:
#include <bits/stdc++.h>
int main() {
  int a, b, c;
  while (scanf("%d%d%d", &a, &b, &c) != EOF) {
    int x, y, z;
    scanf("%d%d%d", &x, &y, &z);
    int f1 = a - x;
    int f2 = b - y;
    int f3 = c - z;
    int need = 0;
    int gong = 0;
    if (f1 < 0)
      need += -2 * (f1);
    else
      gong += f1;
    if (f2 < 0)
      need += -2 * (f2);
    else
      gong += f2;
    if (f3 < 0)
      need += -2 * (f3);
    else
      gong += f3;
    if (gong < need) {
      printf("No\n");
      continue;
    }
    if (f1 > 0 && f1 % 2) gong -= 1;
    if (f2 > 0 && f2 % 2) gong -= 1;
    if (f3 > 0 && f3 % 2) gong -= 1;
    if (gong >= (need)) {
      printf("Yes\n");
      continue;
    } else {
      printf("No\n");
      continue;
    }
  }
  return 0;
}