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>
#define rep(i, a, b) for(int i = a; i < b; i++)
#define Rep(i, a, b) for(int i = a; i <= b; i++)
#define _GLIBCXX_DEBUG
#define Vl vector<ll>
#define Vs vector<string>
#define ll long long
#define ALL(v) (v).begin(),(v).end()
using namespace std;

const double pie = acos(-1.0);
const ll MOD = 1e9 + 7;

// 最大公約数
ll gcd(ll a, ll b)
{
   if (a%b == 0) {
       return(b);
   } else {
       return(gcd(b, a%b));
   }
}

// 最小公倍数
ll lcm(ll a, ll b){
   return a / gcd(a, b) * b;
}

//素数判定
bool is_prime(const unsigned n){
    switch(n){
        case 0: // fall-through
        case 1: return false;
        case 2: // fall-through
        case 3: return true;
    } // n > 3 が保証された
    if(n % 2 == 0 || n % 3 == 0) return false;
    for(unsigned i = 5; i * i <= n; i += 6) {
        if(n % i     == 0) return false; // (6の倍数)-1
        if(n % (i+2) == 0) return false; // (6の倍数)+1
    }
    return true;
}

void solve() {
    ll n; cin >> n;
    ll c1 = 0, c2 = 0, c4 = 0;
    rep(i, 0, n){
        ll a; cin >> a;
        if(a == 2) c2++;
        else if(a%2 == 1) c1++;
        else if(a%4 == 0) c4++;
    }
    // cout << c1 << " " << c2 << " " << c4 << endl;
    if(c2 == 0){
        if(c4+1 >= c1) cout << "Yes" << endl;
        else cout << "No" << endl;
    }else{
        if(c4 >= c1) cout << "Yes" << endl;
        else cout << "No" << endl;
    }
}

int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);

    solve();
    return 0;
}