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>

int main(void){
    int N;
    scanf("%d",&N);
    int X[2][N];
    for(int i=0;i<2;i++){
        for(int j=0;j<N;j++){
            scanf("%d",&X[i][j]);
        }
    }
    int diff = 0;
    int pool = 0;
    for(int i=0;i<N-1;i++){
        diff = X[0][i]-X[1][i];
        if(diff>0){
            X[0][i+1] += 2*diff;
        }else{
            if(diff%2 !=0){
                X[1][i+1] += (-1*diff+1)/2 -1;
            }else{
                X[1][i+1] += -1*diff/2;
            }
        }
    }
    if(X[0][N-1]<=X[1][N-1]){
        printf("Yes");
    }else{
        printf("No");
    }
    /*
    int sumA=0,sumB=0;
    for(int i=0;i<N;i++){
        sumA += X[0][i];
        sumB += X[1][i];
    }
    int count = sumB-sumA;
    if(count<0){
        printf("No");
        return 0;
    }
    int countA=0,countB = 0;
    int diff=0;
    for(int i=0;i<N;i++){
        if(X[0][i]>X[1][i]){
            countB += X[0][i]-X[1][i];
        }else if(X[0][i]<X[1][i]){
            diff = X[1][i]-X[0][i];
            if(diff%2!=0){
                countA += diff+1;
                countB += 1;
            }else{
                countA += diff;
            }
        }
    }
    //printf("%d,%d,%d\n",count,countA,countB);
    if(count>=countB&&2*count-countA==count-countB){
        printf("Yes");
    }else{
        printf("No");
    }
    */
    return 0;
}