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++ (GCC 9.2.1) (judge verdict: Accepted)

#include<cstdio>
#include<cstring>
#include<cmath>
#include<cassert>
#include<iostream>
#include<iomanip>
#include<map>
#include<unordered_map>
#include<set>
#include<unordered_set>
#include<vector>
#include<algorithm>
#include<string>
#include<queue>
using namespace std;
#define DEBUG(x) cout<<#x<<"="<<x<<endl
#define DEBUG2(x,y) cout<<#x<<"="<<x<<","<<#y<<"="<<y<<endl
typedef long long ll;
// #define LOCAL
const int MAXN=1;
const ll MOD=1;
ll _gcd(ll a,ll b){if(b==0)return a;return _gcd(b,a%b);}
ll gcd(ll a,ll b){a=abs(a),b=abs(b);if(a<b)swap(a,b);return _gcd(a,b);}
ll qpow(ll a,ll n){ll rt=1;while(n){if(n&1)rt=(rt*a)%MOD;a=a*a%MOD;n>>=1;}return rt;}
ll factor[MAXN];
void cal_factor(){factor[0]=1;for(int u=1;u<MAXN;u++){factor[u]=(factor[u-1]*u)%MOD;}}
ll C(ll n,ll k){return factor[n]*qpow(factor[n-k],MOD-2)%MOD*qpow(factor[k],MOD-2)%MOD;}
int N;
void solve()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cin>>N;
    int c0=0,c1=0,c2=0;
    for(int u=0;u<N;u++){
        int cnt=0;
        int a;
        cin>>a;
        while(a%2==0){
            a/=2;
            cnt++;
        }
        if(cnt==0)c0++;
        else if(cnt==1)c1++;
        else c2++;
    }
    if(c1==0){
        if(c0-c2<=1)cout<<"Yes\n";
        else cout<<"No\n";
    }
    else {
        if(c0<=c2)cout<<"Yes\n";
        else cout<<"No\n";
    }
}
int main()
{
#ifdef LOCAL
    freopen("in.txt","r",stdin);
#endif
    solve();
}