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,MAX = 510000;
long int pow107 = (int)pow(10, 9) + 7,llimit = LONG_MAX,fac[MAX], finv[MAX], inv[MAX];
int ilimit = INT_MAX;

// テーブルを作る前処理
void COMinit() {
    fac[0] = fac[1] = 1;
    finv[0] = finv[1] = 1;
    inv[1] = 1;
    for (int i = 2; i < MAX; i++){
        fac[i] = fac[i - 1] * i % MOD;
        inv[i] = MOD - inv[MOD%i] * (MOD / i) % MOD;
        finv[i] = finv[i - 1] * inv[i] % MOD;
    }
}

// 二項係数計算
long int COM(long int n,long int k){
    if (n < k) return 0;
    if (n < 0 || k < 0) return 0;
    return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
}

bool comp(pair<int, int> a, pair<int, int> b){
    if(a.first != b.first) return a.first > b.first;
    else return a.second < b.second;
}

long int jou(long int x, long int y){
    long int f = 1;
    for(int i = 0; i < y; ++i){
        f = f * x % pow107;
    }
    return f;
}

bool pn(long int x){
    if(x != 2 && x % 2 == 0)
        return false;
    for(int i = 3; i < x; ++i){
        if(x != i && x % i == 0)
            return false;
    }
    return true;
}

long int factorial(long int n){
    long int f = 1;
    for(long int i = 1; i <= n; ++i){
        f = f * i % pow107;
    }
    return f;
}

long int Func2(long int n,long int m){
    long int s2 = 1,
        r = 1;
    for(long int i = 1; i <= m; ++i){
        s2 = (s2*(n - (i - 1))/i) % (pow107);
    }
    return s2;
}

bool secsort(pair<int, int> a, pair<int, int> b){
    if(a.second == b.second)return a.first > b.first;
    else return a.second > b.second;
}

//long intの最大llimit
//intの最大ilimit
//10^9+7    pow107
//素数pn
//n乗jou
//階乗factorial
//コンビネーションFunc2
//コンビネーションCOMinit(),COM(n,m)

int main(){
    long int n,cc[3] = {};
    cin >> n;
    for(int i = 0 ; i < n ; i++){
        long int a,c = 0;
        cin >> a;
        while(a % 2 == 0 && c != 2){
            c++;
            a/=2;
            if(c == 2)
                break;
        }
        cc[c]++;
        //cout << c << endl;
    }
    //cout << cc[0] << " " << cc[1] << " " << cc[2] << endl;
    if((cc[1] == 0 && cc[0] <= cc[2] + 1) || (cc[1] != 0 && cc[1] % 2 == 0 && cc[0] <= cc[1]/2 + cc[2]) || (cc[1] != 0 && cc[1] % 2 == 1 && cc[0] + cc[1] - cc[1]/2 <= cc[1]/2 + cc[2] + 1)){
        cout << "Yes" << endl;
        return 0;
    }
    cout << "No" << endl;

}