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<cstdlib>
#include<cstring>
#include<cmath>
#include<ctime>
#include<cctype>
#include<climits>
#include<iostream>
#include<string>
#include<vector>
#include<map>
#include<list>
#include<queue>
#include<deque>
#include<algorithm>
#include<numeric>
#include<utility>
#include<complex>
#include<memory>
#include<functional>
 
using namespace std;
 
#define ALL(g) (g).begin(),(g).end()
#define REP(i, x, n) for(int i = x; i < n; i++)
#define rep(i,n) REP(i,0,n)
#define F(i,j,k) fill(i[0],i[0]+j*j,k)
#define P(p) cout<<(p)<<endl;
#define EXIST(s,e) ((s).find(e)!=(s).end())
#define INF 1<<25
#define pb push_back
 
typedef vector<int> vi;
typedef vector<long long> vl;
typedef vector<double> vd;
typedef pair<int,int> pii;
typedef pair<long,long> pll;
typedef long long ll;
int a[100001];
int main(){
	int n,t_2=0,t_4=0;
	cin>>n;
	rep(i,n){
		cin>>a[i];
		if(a[i]%4==0){
			t_4++;
			continue;
		}
		if(a[i]%2==0)
			t_2++;
	}
	int ne=n,ne_2;
	ne-=t_4;
	ne_2=t_2/2;
	ne-=ne_2;

	int nx=n;
	nx-=t_4;
	nx-=t_2;
	if(nx<=t_4&&(t_2!=1)){
		cout<<"Yes"<<endl;
		return 0;
	}
	if(ne<=t_4*2)
		cout<<"Yes"<<endl;
	else 
		cout<<"No"<<endl;



	return 0;
}