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<bits/stdc++.h>
using namespace std;
#define rp(i,n) for(int i=0;i<n;i++)
#define rep(i,m,n) for(int i=m;i<=n;i++)
#define lrp(i,n) for(long long i=0;i<n;i++)
#define lrep(i,m,n) for(long long i=m;i<=n;i++)
#define ll long long
#define sort(a) sort(a.begin(),a.end())
#define reverse(a) reverse(a.begin(),a.end())
#define vi vector<int>
#define vl vector<long long>
#define pb push_back
#define abs(a,b) (a-b>0?a-b:b-a)
#define size size()

int main(){
    int n;
    cin >> n;
    vi a(n);
    rp(i,n){
        cin >> a.at(i);
    }
    vi b(n);
    rp(i,n){
        b.at(i)=0;
        while(a.at(i)%2==0){
            a.at(i)/=2;
            b.at(i)++;
        }
    }
    sort(b);
    int x=0,y=0,z=0;
    rp(i,n){
        if(b.at(i)==0){
            x++;
        }else if(b.at(i)==1){
            y++;
        }else if(b.at(i)>=2){
            z++;
        }
    }
    z -= (x%2?x-1:x);
    if(z<0){
        cout << "No" << endl;
    }else if(y>=1&&z>=1){
        cout << "Yes" << endl;
    }else if(y==0){
        cout << "Yes" << endl;
    }else if(y>=1&&z==0){
        cout << "No" << endl;
    }
    
    return 0;
}