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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.Scanner;

public class Main {

	public static void main(String[] args) {
		no2();
	}

	static void no2() {
		try (Scanner sc = new Scanner(System.in)) {
			int N = sc.nextInt();
			long[] a = new long[N];
			long[] b = new long[N];

			long[] d = new long[N];
			long[] dc = new long[N];

			long asum = 0;
			long bsum = 0;

			long plus1 = 0;
			long plus2 = 0;
			long minus = 0;

			for (int i = 0; i < N; i++) {
				a[i] = sc.nextLong();
				asum += a[i];
			}

			for (int i = 0; i < N; i++) {
				b[i] = sc.nextLong();
				bsum += b[i];

				long t = b[i] - a[i];
				if (t <= 0) {
					minus -= t;
				} else {
					if (t % 2 == 0) {
						plus1 += t;
					} else {
						plus1 += t - 1;
						plus2++;
					}
				}
			}

			long t = plus1 - (minus * 2);
			long tt = minus + plus2 + t;

			if (t < 0) {
				System.out.println("No");
				return;
			}

			if (tt > bsum - asum) {
				System.out.println("No");
			} else {
				System.out.println("Yes");
			}
		}
	}

	static void no1() {
		try (Scanner sc = new Scanner(System.in)) {
			long x = sc.nextLong();
			long y = sc.nextLong();

			if (x % y == 0) {
				System.out.println("-1");
				return;
			}

			long i = 1;
			long t = x;
			while (true) {
				if (t % y != 0) {
					System.out.println(t);
					return;
				}
				if (t > 1000000000000000000L - x) {
					System.out.println("-1");
					return;
				}
				t += x;
			}
		}
	}

}