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;

typedef long long ll;
typedef long double ld;

const int INF = (int) 1e9 + 123;
const ll LINF = (ll) 1e18 + 123;
const ld EPS = (ld) 1e-12;
const ll MOD = (ll) 1e9 + 7;

#define sz(x) (int) (x).size()
#define mp(x, y) make_pair(x, y)
#define pb push_back
#define all(x) (x).begin(), (x).end()
#define lb(s, t, x) (int) (lower_bound(s, t, x) - s)
#define ub(s, t, x) (int) (upper_bound(s, t, x) - s)
#define rep(i, f, t) for (auto i = f; i < t; i++)
#define per(i, f, t) for (auto i = f; i >= t; i--)

ll power(ll x, ll y, ll mod = MOD) {
    if (y == 0) {
        return 1;
    }
    if (y & 1) {
        return power(x, y - 1, mod) * x % mod;
    } else {
        ll tmp = power(x, y / 2, mod);
        return tmp * tmp % mod;
    }
}

void add(ll &x, ll y, ll mod = MOD) {
    x += y;
    if (x >= mod) x -= mod;
    if (x < 0) x += mod;
}

ll sum(ll x, ll y, ll mod = MOD) {
    add(x, y, mod);
    return x;
}

ll mult(ll x, ll y, ll mod = MOD) {
    return (x * y) % mod;
}

ll div(ll x, ll y, ll mod) {
    return x * power(y, mod - 2, mod);
}

template<typename A, typename B> bool mini(A &x, const B &y) {
    if (y < x) {
        x = y;
        return true;
    }
    return false;
}

template<typename A, typename B> bool maxi(A &x, const B &y) {
    if (y > x) {
        x = y;
        return true;
    }
    return false;
}

template<typename T> void read(T first, T last) {
    while (first != last) {
        cin >> *(first++);
    }
}

template<typename T> void print(T first, T last) {
    while (first != last) {
        cout << *(first++) << " ";
    }
    cout << "\n";
}

void run();

#define TASK ""

int main() {
#ifdef LOCAL
    if (strlen(TASK) > 0) {
        cerr << "Reminder: you are using file i/o, filename: " TASK "!" << endl << endl;
    }
#endif
#ifndef LOCAL
    if (strlen(TASK)) {
        freopen(TASK ".in", "r", stdin);
        freopen(TASK ".out", "w", stdout);
    }
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
#endif
    cout << fixed << setprecision(12);
    run();
    return 0;
}

// == SOLUTION == //

const int N = (int) 1e5 + 123;

int n;
int a[N];
int cnt[3];

void run() {
    cin >> n;
    
    rep(i, 0, n) {
        cin >> a[i];
        if (a[i] % 4 == 0) {
            a[i] = 2;
        } else if (a[i] % 2 == 0) {
            a[i] = 1;
        } else {
            a[i] = 0;
        }
        cnt[a[i]]++;
    }
    
    if (((cnt[0] <= cnt[2] + 1) && (cnt[1] == 0)) || (cnt[0] <= cnt[2])) {
        cout << "Yes\n";
    } else {
        cout << "No\n";
    }
}