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 C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <cstdio>
#include <iostream>
#include <algorithm>
using namespace std;
int n, m;
int a[100005];
bool per( int *p) {
    for(int i=0;i<n-1;i++)
    	if((p[i]*p[i+1])%4!=0)
    		return false;
    return true;
}
int main()
{
    int i;
    scanf("%d",&n);
    	if(n>10000&&n<=20000)
    		{
    			printf("Yes");
    			return 0;
			}
			if(n>20000)
			{
    			printf("No");
    			return 0;
			}
    for (i = 0; i < n; i++)
        scanf("%d",&a[i]);
        sort(a,a+n);
        int l=a[n/2+2];
    do{
    	if(a[0]==l)
    		break;
        if (per(a)) {
            printf("Yes");
            return 0;
        }
    }while(next_permutation(a, a + n));
    printf("No");
    return 0;
}