Problem:
Score : 400 points

Problem Statement

We have a sequence of length N, a = (a_1, a_2, ..., a_N).
Each a_i is a positive integer.

Snuke's objective is to permute the element in a so that the following condition is satisfied:

For each 1 ≤ i ≤ N - 1, the product of a_i and a_{i + 1} is a multiple of 4.

Determine whether Snuke can achieve his objective.

Constraints

2 ≤ N ≤ 10^5

a_i is an integer.

1 ≤ a_i ≤ 10^9

Input

Input is given from Standard Input in the following format:

N
a_1 a_2 ... a_N

Output

If Snuke can achieve his objective, print Yes; otherwise, print No.

Sample Input 1

3
1 10 100

Sample Output 1

Yes

One solution is (1, 100, 10).

Sample Input 2

4
1 2 3 4

Sample Output 2

No

It is impossible to permute a so that the condition is satisfied.

Sample Input 3

3
1 4 1

Sample Output 3

Yes

The condition is already satisfied initially.

Sample Input 4

2
1 1

Sample Output 4

No

Sample Input 5

6
2 7 1 8 2 8

Sample Output 5

Yes

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)


import java.util.Scanner;

public class Main {

	public static void main(String[] args) {
		int[] a = null;
		try(Scanner sc = new Scanner(System.in)){
			// N入力 aスペース区切り入力
			a = new int[sc.nextInt()];
			for(int i = 0; i < a.length; i ++) {
				a[i] = sc.nextInt();
			}
		}
		
		// a要素の4の倍数カウント
		int cnt4 = 0;
		// a要素の2の倍数カウント
		int cnt2 = 0;
		for(int i = 0; i < a.length; i++) {
			
			// 要素が4の倍数
			if(a[i] % 4 == 0) {
				cnt4++;
			}
			// 要素が2の倍数
			else if(a[i] % 2 == 0){
				cnt2++;
			}
		}
		// 2の倍数の数を要素数から排除
		int n = a.length - cnt2;
		
		// 2の倍数が存在した
		if(cnt2 != 0){
			// 偶数
			if(n % 2 == 0) {
				// 4の倍数が要素の半分以上
				if((n /2) <= cnt4) {
					System.out.println("Yes");
				} else {
					System.out.println("No");
				}				
			} else {
				// 4の倍数が要素の半分より上
				if((n /2) < cnt4) {
					System.out.println("Yes");
				} else {
					System.out.println("No");
				}				
			}
		} else {
			// 4の倍数が要素の半分以上
			if((n /2) <= cnt4) {
				System.out.println("Yes");
			} else {
				System.out.println("No");
			}
		}
	}
}