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 C (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <limits.h>
#include <stdint.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <time.h>
#include <unistd.h>
#include <math.h>

#define P 1000000007

int comp(const void *a, const void *b){return *(int*)a-*(int*)b;}

int main(void){
	int n, i;
	int a[10000], b[10000];
	long A=0, B=0;
	scanf("%d", &n);
	for(i=0; i<n; i++){ scanf("%d", &a[i]); A+=a[i]; }
	for(i=0; i<n; i++){ scanf("%d", &b[i]); B+=b[i]; }
	int c=0, d=0;
	for(i=0; i<n; i++){
		if(a[i]<b[i]){c+=(b[i]-a[i]+1)/2; d+=(b[i]-a[i]+1)%2;}
		if(a[i]>b[i]){d+=a[i]-b[i];}
	}
	puts(c>=d?"Yes":"No");
}