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>
using namespace std;
template <class T>
inline void umax(T &a, T b) {
  if (a < b) a = b;
}
template <class T>
inline void umin(T &a, T b) {
  if (a > b) a = b;
}
template <class T>
inline T abs(T a) {
  return a > 0 ? a : -a;
}
template <class T>
inline T gcd(T a, T b) {
  return __gcd(a, b);
}
template <class T>
inline T lcm(T a, T b) {
  return a / gcd(a, b) * b;
}
long long modpow(long long a, long long n, long long temp) {
  long long res = 1, y = a;
  while (n > 0) {
    if (n & 1) res = (res * y) % temp;
    y = (y * y) % temp;
    n /= 2;
  }
  return res % temp;
}
long long ison(long long mask, long long pos) { return (mask & (1 << pos)); }
long long cbit(long long n) {
  long long k = 0;
  while (n) n &= (n - 1), k++;
  return k;
}
long long nbit(long long n) {
  long long k = 0;
  while (n) n /= 2, k++;
  return k;
}
long long mod = 1e9 + 7;
int sgn(long long x) { return x < 0 ? -1 : !!x; }
long long xo(long long i) {
  if ((i & 3) == 0) return i;
  if ((i & 3) == 1) return 1;
  if ((i & 3) == 2) return i + 1;
  return 0;
}
int main() {
  long long a, b, c, x, y, z;
  cin >> a >> b >> c;
  cin >> x >> y >> z;
  long long ex = 0;
  if (a > x) {
    ex += (a - x) / 2;
    a = x;
  }
  if (b > y) {
    ex += (b - y) / 2;
    b = y;
  }
  if (c > z) {
    ex += (c - z) / 2;
    c = z;
  }
  if (x > a) {
    if (ex >= (x - a)) {
      ex -= (x - a);
      a = x;
    }
  }
  if (y > b) {
    if (ex >= (y - b)) {
      ex -= (y - b);
      b = y;
    }
  }
  if (z > c) {
    if (ex >= (z - c)) {
      ex -= (z - c);
      c = z;
    }
  }
  if (a >= x && b >= y && c >= z)
    cout << "YES";
  else
    cout << "NO";
}