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;

    const int MOD = 1000000007;
    const double PI = acos(-1); 

    //long long
    using ll = long long;

    //出力系
    #define print(x) cout << x << endl
    #define yes cout << "Yes" << endl
    #define YES cout << "YES" << endl
    #define no cout << "No" << endl
    #define NO cout << "NO" << endl
    
    // begin() end()
    #define all(x) (x).begin(),(x).end()

    //for
    #define REP(i,n) for(int i=0, i##_len=(n); i<i##_len; ++i)

    //最大公約数（ll）
    ll gcd(ll x, ll y) { return y ? gcd(y,x%y) : x;}

    //素因数分解
    vector<pair<ll, int>>factorize(ll n){
      vector<pair<ll, int>> res;
      for(ll i = 2; i * i <= n; ++i){
        if(n % i) continue;
        res.emplace_back(i, 0);
        while(n % i == 0){
          n /= i;
          res.back().second++;
        }
      }
    if(n != 1) res.emplace_back(n, 1);
    return res;
    }



  // // 二分探索(最小値を出す問題は疑う)
  // bool C(ll x){
  //     ll res = 0;
  //     for(int i = 0; i < N; ++i){
  // 		res += max(0ll, A[i]-x/F[i]);
  //     }
  //     return res <= K;
  // }

  //   ll binary_search(ll ok, ll ng){
  //     while(abs(ok - ng) > 1){
  //       ll mid = (ok + ng) / 2;
  //       if(C(mid)) ok = mid;
  //       else ng = mid;
  //     }
  //     return ok;
  //   }


int main(){
int N;
cin >> N;
vector<ll>a(N);
REP(i, N) cin >> a.at(i);
 
ll four = 0, odd = 0, two = 0;
 
REP(i, N){
	if(a.at(i) % 4 == 0 ) four++;
	else if(a.at(i) % 4 != 0 && a.at(i) % 2 == 0) two = 1;
	else if(a.at(i) % 2 == 1) odd++;
}
// print(odd);
// print(four);
 
if(odd == 0) yes;
else if(four >= odd + two - 1) yes;
else no;
}