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: Runtime Error)

#include <stdio.h>
#include <algorithm>
using namespace std;
static bool flag=false;
void Swap(int *lhs, int *rhs)
{
	int t = *lhs;

	*lhs = *rhs;
	*rhs = t;
}

/************************************************************************/
/* 功能：实现全排列功能
/* 参数：
/*       source--整数数组，存放需要全排列的元素
/*       begin --查找一个排列的开始位置
/*       end   --查找一个排列的结束位置，当begin=end时，表明完成一个排列
/************************************************************************/
void FullPermutation(int source[], int begin, int end)
{
	int i;
if(flag==true)
return;
	if (begin >= end) // 找到一个排列
	{
		/*for (i = 0; i < end; i++)
		{
			printf("%d", source[i]);
		}
		printf("\n");*/
		for (i = 0; i < end-1; i++)
		{
			if ((source[i] * source[i + 1]) % 4 != 0)
				break;
		}
		if (i == end - 1)
		{
			flag = true;
		}
		
	}
	else// 没有找完一个排列，则继续往下找下一个元素
	{
		for (i = begin; i < end; i++)
		{
			if (begin != i)
			{
				Swap(&source[begin], &source[i]); // 交换
			}

			// 递归排列剩余的从begin+1到end的元素
			FullPermutation(source, begin + 1, end);

			if (begin != i)
			{
				Swap(&source[begin], &source[i]); // 回溯时还原
			}
		}
	}
}

int main()
{
	int source[30];
	int i, count;
	
	scanf("%d", &count);

	// 初始化数组
	for (i = 0; i < count; i++)
	{
		scanf("%d", &source[i]);
	}

	FullPermutation(source, 0, count);
	if (flag == true)
		printf("Yes");
	else
		printf("No");
	return 0;
}