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>
#include <stdlib.h>

int main() {
	int n;
	int a[10240], b[10240];
	int i, j, t, d;
	scanf("%d", &n);

	for (i = 0; i < n; i ++) {
		scanf("%d ", &a[i]);
	}

	for (i = 0; i < n; i ++) {
		scanf("%d ", &b[i]);
	}

	for (i = 0; i < n-1; i++) {
		for (j = n-1; j > i; j --) {
			if (a[j-1] < a[j]) {
				t = a[j-1];
				a[j-1] = a[j];
				a[j] = t;
			}
		}
	}

	for (i = 0; i < n-1; i++) {
		for (j = n-1; j > i; j --) {
			if (b[j-1] < b[j]) {
				t = b[j-1];
				b[j-1] = b[j];
				b[j] = t;
			}
		}
	}

	for (i = 0; i < n - 1; i ++) {
		if (b[i] - a[i] > (a[i+1] - b[i+1]) * 2) {
			a[i] += (a[i+1] - b[i+1]) * 2;
			b[i+1] += (a[i+1] - b[i+1]);
		}
	}
	d = 1;
	for (i = 0; i < n; i ++) {
		if (a[i] > b[i]) {
			d = 0;
		}
	}

	printf ("%s", (d==1 ? "Yes" : "No"));

	/*
	printf("%d\n", n);
	for (i = 0; i < n; i ++) {
		printf("%d ", a[i]);
	}
	printf("\n", a[i]);
	for (i = 0; i < n; i ++) {
		printf("%d ", b[i]);
	}
	printf("\n", a[i]);
	*/
	return 0;
}