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 PYTHON3):

Solution:
''' ===============================
    -- @uthor : Kaleab Asfaw
    -- Handle : kaleabasfaw2010
    -- Bio    : High-School Student
    ==============================='''

# Fast IO
import sys
import os
from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
    newlines = 0
    def __init__(self, file): self._fd = file.fileno(); self.buffer = BytesIO(); self.writable = "x" in file.mode or "r" not in file.mode; self.write = self.buffer.write if self.writable else None
    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            if not b: break
            ptr = self.buffer.tell(); self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0; return self.buffer.read()
    def readline(self):
        while self.newlines == 0:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE)); self.newlines = b.count(b"\n") + (not b); ptr = self.buffer.tell(); self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines -= 1; return self.buffer.readline()
    def flush(self):
        if self.writable: os.write(self._fd, self.buffer.getvalue()); self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
    def __init__(self, file): self.buffer = FastIO(file); self.flush = self.buffer.flush; self.writable = self.buffer.writable; self.write = lambda s: self.buffer.write(s.encode("ascii")); self.read = lambda: self.buffer.read().decode("ascii"); self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout); input = lambda: sys.stdin.readline().rstrip("\r\n")
# Others
# from math import floor, ceil, gcd
# from decimal import Decimal as d
mod = 10**9+7
def lcm(x, y): return (x * y) / (gcd(x, y))
def fact(x, mod=mod):
    ans = 1
    for i in range(1, x+1): ans = (ans * i) % mod
    return ans
def arr2D(n, m, default=0): return [[default for j in range(m)] for i in range(n)]
def arr3D(n, m, r, default=0): return [[[default for k in range(r)] for j in range(m)] for i in range(n)]
def sortDictV(x): return {k: v for k, v in sorted(x.items(), key = lambda item : item[1])}
class DSU:
    def __init__(self, length): self.length = length; self.parent = [-1] * self.length  # O(log(n))
    def getParent(self, node, start):  # O(log(n))
        if node >= self.length: return False
        if self.parent[node] < 0:
            if start != node: self.parent[start] = node
            return node
        return self.getParent(self.parent[node], start)
    def union(self, node1, node2):  # O(log(n))
        parent1 = self.getParent(node1, node1); parent2 = self.getParent(node2, node2)
        if parent1 == parent2: return False
        elif self.parent[parent1] <= self.parent[parent2]: self.parent[parent1] += self.parent[parent2]; self.parent[parent2] = parent1
        else: self.parent[parent2] += self.parent[parent1]; self.parent[parent1] = parent2
        return True
    def getCount(self, node):  return -self.parent[self.getParent(node, node)] # O(log(n))

from math import ceil
a, b, c = list(map(int, input().split()))
d, e, f = list(map(int, input().split()))

extra = max(0, (a-d)//2)
extra += max(0, (b-e)//2)
extra += max(0, (c-f)//2)

needed = max(0, (d-a))
needed += max(0, (e-b))
needed += max(0, (f-c))

if extra < needed:
    print("No")
else:
    print("Yes")

# diff = abs(a - d) + abs(b - e) + abs(c - f)
# diff = ceil(diff / 4)
# if a + b + c - diff >= d + e + f:
#     print("Yes")
# else:
#     print("No")