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<stdio.h>

#define MAX 1000000000
#define NMAX 10000

int main(void){
    int N,a[NMAX],b[NMAX],sum_a,sum_b,i,j,red_ab;
    scanf("%d", &N);
    for(i = 0; i < N; i++){
        scanf("%d", &a[i]);
    }
    for(i = 0; i < N; i++){
        scanf("%d", &b[i]);
    }
    sum_a = 0;
    sum_b = 0;    
    for(i = 0; i < N; i++){
        sum_a += a[i];
        sum_b += b[i];
    }
    red_ab = sum_b - sum_a;
    if(red_ab < 0){
        printf("No");
        return 0;
    }
    else{
        j = 0;
        for(i = 0; i < N; i++){
            if(a[i] > b[i]){
                j += a[i] - b[i];
            }
            else if(b[i] > a[i]){
                j += (b[i] - a[i]) / 2 + (b[i] - a[i]) % 2;                
            }
            if(j >= red_ab){
                printf("No");
                return 0;
            }
        }
    }
    printf("Yes");
    return 0;
}