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 <iostream>
#include <iomanip>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <climits>
#include <complex>
#include <vector>
#include <list>
#include <set>
#include <queue>
#include <stack>
#include <map>
#include <string>
#include <algorithm>
#include <numeric>
using namespace std;

#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define FOR(i, a, b) for(ll i = static_cast<ll>(a); i < static_cast<ll>(b); i++)
#define FORR(i, a, b) for(ll i = static_cast<ll>(a); i >= static_cast<ll>(b); i--)
#define REP(i, n) for(ll i = 0ll; i < static_cast<ll>(n); i++)
#define REPR(i, n) for(ll i = static_cast<ll>(n); i >= 0ll; i--)
#define ALL(x) (x).begin(), (x).end()

typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> P;
typedef pair<ll, ll> LP;
typedef pair<int, P> IP;
typedef pair<ll, LP> LLP;

const int dx[] = {1, -1, 0, 0};
const int dy[] = {0, 0, 1, -1};

constexpr int INF = 100000000;
constexpr ll LINF = 10000000000000000ll;
constexpr int MOD = static_cast<int>(1e9 + 7);
constexpr double EPS = 1e-9;
// }}}

int N;
ll a[100000];

void init()
{
}

void solve()
{
    int num[3] = {0, 0, 0};
    REP(i, N){
        if(a[i] % 4 == 0) num[0]++;
        else if(a[i] % 2 == 0) num[1]++;
        else num[2]++;
    }
    cout << ((num[1] == 0 && num[0]+1 >= num[2]) || (num[1] > 0 && num[0] >= num[2]) ?
            "Yes" : "No") << endl;
}

int main()
{
    cin.tie(0);
    ios::sync_with_stdio(false);
    cin >> N;
    REP(i, N) cin >> a[i];
    solve();
    return 0;
}

// vim:set foldmethod=marker: