Problem:
Score : 300 points

Problem Statement

You are given two integer sequences of length N: a_1,a_2,..,a_N and b_1,b_2,..,b_N.
Determine if we can repeat the following operation zero or more times so that the sequences a and b become equal.

Operation: Choose two integers i and j (possibly the same) between 1 and N (inclusive), then perform the following two actions simultaneously:

Add 2 to a_i.

Add 1 to b_j.

Constraints

1 ≤ N ≤ 10 000

0 ≤ a_i,b_i ≤ 10^9 (1 ≤ i ≤ N)

All input values are integers.

Input

Input is given from Standard Input in the following format:

N
a_1 a_2 .. a_N
b_1 b_2 .. b_N

Output

If we can repeat the operation zero or more times so that the sequences a and b become equal, print Yes; otherwise, print No.

Sample Input 1

3
1 2 3
5 2 2

Sample Output 1

Yes

For example, we can perform three operations as follows to do our job:

First operation: i=1 and j=2. Now we have a = \{3,2,3\}, b = \{5,3,2\}.

Second operation: i=1 and j=2. Now we have a = \{5,2,3\}, b = \{5,4,2\}.

Third operation: i=2 and j=3. Now we have a = \{5,4,3\}, b = \{5,4,3\}.

Sample Input 2

5
3 1 4 1 5
2 7 1 8 2

Sample Output 2

No

Sample Input 3

5
2 7 1 8 2
3 1 4 1 5

Sample Output 3

No

Solution in Python (3.8.2) (judge verdict: Accepted)

#from statistics import median
#import collections
#aa = collections.Counter(a) # list to list || .most_common(2)で最大の2個とりだせるお a[0][0]
from fractions import gcd
from itertools import combinations,permutations,accumulate # (string,3) 3回
#from collections import deque
from collections import deque,defaultdict,Counter
import decimal
import re
#import bisect
#
#    d = m - k[i] - k[j]
#    if kk[bisect.bisect_right(kk,d) - 1] == d:
#
#
#
# pythonで無理なときは、pypyでやると正解するかも！！
#
#
# my_round_int = lambda x:np.round((x*2 + 1)//2)
# 四捨五入
import sys
sys.setrecursionlimit(10000000)
mod = 10**9 + 7
#mod = 9982443453
def readInts():
  return list(map(int,input().split()))
def I():
  return int(input())
n = I()
A = readInts()
B = readInts()
biga = 0
bigb = 0
for i in range(n):
    if A[i] > B[i]:
        biga += A[i]-B[i]
    elif A[i] < B[i]:
        bigb += (B[i]-A[i])//2
print('Yes' if bigb >= biga else 'No')