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 <iostream>
#include <string>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <numeric>
#include <regex>
#include <climits>

#define FOR(i, f, n) for(int (i) = (f); (i) < (n); (i)++)
#define RFOR(i, f, n) for(int (i) = (f); (i) >= (n); (i)--)
#define REP(i, n) for(int (i) = 0; (i) < (n); (i)++)
#define RREP(i, n) for(int (i) = (n); (i) >= 0; (i)--)
#define rep(i, n) for(int (i) = 0; (i) < (n); (i)++)
#define rrep(i, n) for(int (i) = (n); (i) >= 0; (i)--)
#define forin(i, a) for(auto i: (a))
#define p(...) printf(__VA_ARGS__); cout << endl;
#define in(...) __VA_ARGS__; input__(__VA_ARGS__);
#define input(...) in(__VA_ARGS__)
#define out(...) output__(__VA_ARGS__);
#define output(...) out(__VA_ARGS__)
#define list_input(x, n) x[(n)]; for(int (i) = 0; (i) < (n); (i)++) cin >> x[i];
#define lambda(a, b, c) [](auto const &a, auto const &b){ return c ;}
#define key(t, x, compare) [](const t &a, const t &b){ return a x compare b x ;}
#define unless(a) if(!(a))

using namespace std;

using pii = pair<int, int>;

template <class... T>
void input__(T &...a) {
  for(auto v: {&a...}) {
    cin >> *v;
  }
}

void output__(){
  cout << endl;
}

template <class Head, class... T>
void output__(Head const &v, T const &...a) {
  if(sizeof...(a) == 0) {
    cout << v;
  } else {
    cout << v << " ";
  }
  output__(a...);
}

int main(){
  long in(n);
  long list_input(a, n);

  int n4 = 0;
  int n2 = 0;
  int n1 = n;
  rep(i, n) {
    if(a[i] % 4 == 0) {
      n4++;
    } else if(a[i] % 2 == 0) {
      n2++;
    }
  }

  if(n4 >= (n1 - n4) * 2) {
    out("Yes");
  } else if(n2 % 2 == 0) {
    out("Yes");
  } else if(n4 - (n1 - n4 - n2) / 2 >= 1) {
    out("Yes");
  } else {
    out("No");
  }
}