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()
{
	long long a[10001],b[10002],i,x,n,y;
	while(scanf("%lld",&n)!=EOF)
	{
		x=y=0;
		for(i=0;i<n;i++)
		{
			scanf("%lld",&a[i]);
			x+=a[i];
		}
		for(i=0;i<n;i++)
		{
			scanf("%lld",&b[i]);
			y+=b[i];
		}
		
		if(x>y)
		printf("No\n");
		else if(x==y)
		{
			int k=0;
			for(i=0;i<n;i++)
			{
				if(a[i]!=b[i])
				{
					k=1;break;
				}
			}
			if(k==1)
			printf("Yes\n");
			else
			printf("No\n");
		}
		else
		{
			long long c=0,k,m=0;
			for(i=0;i<n;i++)
			{
				if(b[i]<a[i])
				c+=(a[i]-b[i]);
			}
			
			if(c>(y-x))
			printf("No\n");
			else if(c==(y-x))
			{
				for(i=0;i<n;i++)
				{
					if(b[i]>a[i])
					{
						k=b[i]-a[i];
						if(k%2!=0)
						{
							m=1;
							break;
						}
					}
				}
				if(m==0)
				printf("Yes\n");
				else
				printf("No\n");
			}
			else
			{
				k=y-x-c;
				c=0;
				for(i=0;i<n;i++)
				{
					if(b[i]>a[i])
					{
						c=b[i]-a[i];
						if(c%2!=0)
						m++;
					}
				}
				if(m>k)
				printf("No\n");
				else
				printf("Yes\n");
			}
		}
	}
	return 0;
}