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: Accepted)

#include <bits/stdc++.h>
using namespace std;
#define _GLIBCXX_DEBUG
#define rep(i, n) for (ll i = 0; i < (ll)(n); i++)
using ll = long long;
using ld = long double;
#define rep2(i, s, n) for (ll i = (s); i < (ll)(n); i++)
#define rep3(i, n) for (ll i = 1; i < (ll)(n+1); i++)
#define rep4(i, s, n) for (ll i = (s); i < (ll)(n+1); i++)
#define repr(i,n) for (ll i = (n-1); i>=0;i--)
#define repr3(i,n) for(ll i = (n);i>0;i--)
#define stlen(s) ll s.size()-1
#define all(v) v.begin(), v.end()
#define cout(n) cout<<std::fixed<<std::setprecision(n)
using Graph = vector<vector<int>>;
#define INF1  INT_MAX;
#define INF2  LLONG_MAX;
#define PI  3.14159265358979323846;

bool is_prime(long long N) {
    if (N == 1) return false;
    for (long long i = 2; i * i <= N; ++i) {
        if (N % i == 0) return false;
    }
    return true;
}

int main() {
    int N;
    cin>>N;
    ll a[N];
    rep(i,N){
        cin>>a[i];
    }
    int count0=0,count2=0,count1=0;
    rep(i,N){
        if(a[i]%2!=0){
            count0++;
        }
        else if(a[i]%4==0){
            count2++;
        }
        else if(a[i]%2==0){
            count1++;
        }

}
if(count1>0){
if(count0<=count2){
    cout<<"Yes"<<endl;
}
else{
    cout<<"No"<<endl;
}
}
else{
    if(count0<=count2+1){
        cout<<"Yes"<<endl;
    }
    else{
        cout<<"No"<<endl;
    }
}

    
return 0;
}