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 <iostream>
#include <fstream>
#include <sstream>
#include <set>
#include <map>
#include <vector>
#include <list>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <string>
#include <queue>
#include <bitset>     //UWAGA - w czasie kompilacji musi byc znany rozmiar wektora - nie mozna go zmienic
#include <cassert>
#include <iomanip>        //do setprecision
#include <ctime>
#include <complex>
using namespace std;

#define FOR(i,b,e) for(int i=(b);i<(e);++i)
#define FORQ(i,b,e) for(int i=(b);i<=(e);++i)
#define FORD(i,b,e) for(int i=(b)-1;i>=(e);--i)
#define REP(x, n) for(int x = 0; x < (n); ++x)

#define ST first
#define ND second
#define PB push_back
#define MP make_pair
#define LL long long
#define ULL unsigned LL
#define LD long double

const double pi = 3.141592653589793238462643383279502884197169399375105820974944592307816406286208998628034825342;
const int mod=1000000007;

int main(){
	int n,a[100101]={};
	int c4=0,c2=0;
	cin>>n;
	FOR(i,0,n){
		cin>>a[i];
		if(a[i]%4==0){
			a[i]=2;
			c4++;
		}else if(a[i]%2==0){
			a[i]=1;
			c2++;
		}else{
			a[i]=0;
		}
	}
	n-=(c2+c4);
	if(n<=c4)cout<<"Yes"<<endl;
	else cout<<"No"<<endl;

    return 0;
}