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>

#define rep(i, n) for (int i = 0; i < int(n); i++)
#define rrep(i, n) for (int i = int(n) - 1; i >= 0; i--)
#define reps(i, n) for (int i = 1; i <= int(n); i++)
#define rreps(i, n) for (int i = int(n); i >= 1; i--)
#define repi(i, a, b) for (int i = int(a); i < int(b); i++)
#define each(x, y) for (auto &x : y)
#define all(a) (a).begin(), (a).end()
#define bit(b) (1ll << (b))
#define uniq(v) (v).erase(unique(all(v)), (v).end())
#define rsort(v) sort(all(v)); reverse(all(v))

using namespace std;

using i32 = int;
using i64 = long long;
using f80 = long double;
using vi32 = vector<i32>;
using vi64 = vector<i64>;
using vf80 = vector<f80>;
using vstr = vector<string>;

inline void yes() { cout << "Yes" << '\n'; exit(0); }
inline void no() { cout << "No" << '\n'; exit(0); }
inline i64 gcd(i64 a, i64 b) { if (min(a, b) == 0) return max(a, b); if (a % b == 0) return b; return gcd(b, a % b); }
inline i64 lcm(i64 a, i64 b) { return a / gcd(a, b) * b; }
void solve(); int main() { ios::sync_with_stdio(0); cin.tie(0); cout << fixed << setprecision(16); solve(); return 0; }
template<typename T> class pqasc : public priority_queue<T, vector<T>, greater<T>> {};
template<typename T> class pqdesc : public priority_queue<T, vector<T>, less<T>> {};
template<typename T> inline void amax(T &x, T y) { if (x < y) x = y; }
template<typename T> inline void amin(T &x, T y) { if (x > y) x = y; }
template<typename T> inline T power(T x, i64 n) { T r = 1; while (n > 0) { if (n & 1) r *= x; x *= x; n >>= 1; } return r; }
template<typename T> istream& operator>>(istream &is, vector<T> &v) { for (auto &x : v) is >> x; return is; }
template<typename T> ostream& operator<<(ostream &os, const vector<T> &v) { rep(i, v.size()) { if (i) os << ' '; os << v[i]; } return os; }

void solve() {
  int N; cin >> N;
  vi32 a(N); cin >> a;
  int cnt1 = 0, cnt4 = 0;
  rep(i, N) {
    if (a[i] % 2 == 1) cnt1++;
    if (a[i] % 4 == 0) cnt4++;
  }
  if (cnt1 > cnt4 + 1) no();
  if (cnt1 == cnt4 + 1 && N > cnt1 + cnt4) no();
  yes();
}