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;
#define REP(i,n) for(int i=0, i##_len=(n); i<i##_len; ++i)
#define REPR(i,n) for(int i=n; i>-1; --i)
#define ALL(a) (a).begin(),(a).end()
#define FILL(a,n,x); REP(i,(n)){ (a)[i]=(x); }
#define CINA(a,n); REP(i,(n)){ cin >> (a)[i]; }
#define FILL2(a,n,m,x); REP(i,(n)){ REP(j,(m)){(a)[i][j]=(x);} }
#define CINA2(a,n,m); REP(i,(n)){ REP(j,(m)){cin >> (a)[i][j];} }
#define Liny "Yes\n"
#define Linn "No\n"
#define LINY "YES\n"
#define LINN "NO\n"

//cout << setfill('0') << right << setw(4) << 12; // "0012"
int keta(ll x){ if(x<10){return 1;} else{return keta(x/10) + 1;}}
int keta_wa(ll x){ if(x<10){return x;} else{return keta_wa(x/10) + x%10;} }
int ctoi(char c){ return ( (c>='0' && c<='9')? c - '0': 0 );}
int __stoi(string s){ return atoi(s.c_str()); }
ll sum(ll a[],ll N){ return accumulate(a,a+N,0LL);}
ll gcd(ll a,ll b){if(a<b)swap(a,b); return b?gcd(b,a%b):a;}
ll lcm(ll a,ll b){if(a<b){swap(a,b);} return a/gcd(a,b)*b;}
template<class T> void chmax(T& a, T b){ if(a<b){a=b;} }
template<class T> void chmin(T& a, T b){ if(a>b){a=b;} }
const ll MOD = 1e9+7;
#define pii pair<int,int>
#define pll pair<ll,ll>


int main(){
    int N; cin>>N;
    int num_f = 0;
    int num_t = 0;
    int num_r = 0;
    REP(i,N){
        ll a; cin >> a;
        if(a%4==0){ ++num_f; }
        else if(a%2==0){ ++num_t;}
        else{ ++num_r; }
    }
    bool ok;
    if(num_f>=num_r){ ok = true; }
    else{ 
      if(num_t==0 && num_f>=num_r-1){ ok = true;}
      else{ ok = false;}
    }
    cout << (ok?Liny:Linn);
}