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()
{
    int p[10000], q[10000];
    long  long  N, i, j, k, t, m, s1, s2;
    scanf("%lld",&N);
    s1 = s2 = j = t = k = 0;
    for(i=0; i<N; i++)
    {
        scanf("%d",&p[i]);
        s1 += p[i];
    }
    for(i=0; i<N; i++)
    {
        scanf("%d",&q[i]);
        s2 += q[i];
    }

    if(s1>s2)
        printf("No");

    else if(s1==s2)
    {
        for(i=1; i<N; i++)
            if(p[i]!=q[i])
                break;
        if(i<N)
            printf("No");
        else
            printf("Yes");
    }

    else if(s1<s2)
    {
        while(s1<s2)
        {
            s1 += 2;
            s1++;
            if(s1==s2)
                k = 1;
        }
        if(k==1)
        {
           for(i=0;i<N;i++)
            {
              if(p[i]>q[i])
                   j += p[i]-q[i];
              else if(p[i]<q[i])
                   t += q[i]-p[i];
            }
            if(t%(j*2)==0)
                printf("Yes");
            else
                printf("No");
        }
        else
            printf("No");
    }
    return 0;
}