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 <bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef vector<int> vi;
typedef pair<int, int> ii;
typedef pair<ll, int> li;
typedef pair<int, ll> il;
typedef vector<ii> vii;
typedef vector<il> vil;
typedef vector<li> vli;

#define F            first
#define S            second
#define pb(x)        push_back(x)
#define mp(a,b)      make_pair(a,b)
#define mt(a,b,c)    make_tuple(a,b,c)
#define sz           size()
#define all(A)       A.begin(), A.end()
#define mem(a, b)    memset(a, b, sizeof(a))   // all elements of A are b=0 or b=-1 else don't work
#define f0(i,b)      for(ll i=0;i<(b);i++)
#define f1(i,b)      for(ll i=1;i<=(b);i++)
#define f2(i,a,b)    for(ll i=(a);i<=(b);i++)
#define f3(i,b,a)    for(ll i=(b);i>=(a);i--)
#define rep(i,a,b,c) for(ll i=(a);i!=(b);i+=(c))

#define MOD          1000000007 //10^9 + 7
#define PI           acos(-1.0)
#define MAX3(a,b,c)  max(a,max(b,c))
#define MIN3(a,b,c)  min(a,min(b,c))

/*------------------------Bitmask-------------------------------*/
#define bitOn(N,pos)      (N | (1LL<<(pos)))
#define bitOff(N,pos)     (N & (~(1LL<<(pos))))
#define bitCheck(N,pos)   ((bool)(N & (1LL<<(pos))))
#define bitFlip(a,k)      (a^(1LL<<(k)))

#define whilee(i,t)  while(i!=t)
#define  sl(a)       scanf("%lld",&a)
#define  pl(a)       printf("%lld\n",a)

//x = 1e18 means 10^18
//for(auto it =A.begin(); it!=A.end(); it++)
//sort(A.begin(),A.end()); //sort(A, +n, greater<int>());
//binary_search(A.begin(),A.end(),x);
//reverse(A.begin(), A.end());
//cout << *max_element(A.begin(), A.end()); //will have index if A.begin() is minus from it
//cout << *min_element(A.begin(), A.end());
//count(A.begin(), A.end(), X); //counts the occurrences of X
//distance(A.begin(),A.end());  //distance between first to last element
//accumulate(A.begin(), A.end(), 0); //add
//next_permutation(A.begin(), A.end());
//prev_permutation(A.begin(), A.end());
//swap(S[0], T[0]);

//scanf("%lld",&x);
//printf("%lld\n",x);
//printf("%I64d\n", a);
//while (scanf("%lld %lld", &a, &b) == 2)
//printf("Case %lld: %lld\n", t++, a);

/*----------------------- Graph Moves---------------------------*/
// int dx[] = {1,-1,0,0} , dy[] = {0,0,1,-1};  // 4 Direction
// int dx[] = {0,0,1,-1,-1,1,-1,1} , dy[] = {-1,1,0,0,1,1-1,-1};  // 8 Direction // Kings Move
// int dx[] = {1,-1,1,-1,2,2,-2,-2} , dy[] = {2,2,-2,-2,1,-1,1,-1};  // Knight Direction
// int dx[] = {2,-2,1,1,-1,-1} , dy[] = {0,0,1,-1,1,-1};  // Hexagonal Direction

//vector <ll> A;
//deque <ll> A;
//queue <ll> A;
//stack <ll> A;
//list <int> A;
//set <ll> A;
//map <ll,ll> A;
//unordered_map <ll, ll> A;
//pair <ll,ll> A ;
//tuple <ll, ll, ll> A;
//vector <pair <ll,ll>> A;
//bitset<32> bset1;
//bool ok = true;
//char ch;
//string S, T, U;

ll i=0, j=0, k=0, a, b, c, d, e, f, g, h, l, m, n, o, p, q, r, s, t=0, u, v, w, x, y, z, maxx=INT_MIN, minn=INT_MAX, ans=0, sum=0, flag=0, cnt=0;

int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);//NULL
    cout.tie(nullptr);//NULL

    // while(cin>>n){}
    cin>>n;
    ll A[n];

    f0(i,n)
    {
        cin>>x;
        if(x%2==1)
        {
            t++;
        }
        else
        {
            if(x%4==0)
            {
                k++;
            }
            else
            {
                j++;
            }
        }
    }

    if(j==0)
    {
        if(t<=k+1)
        {
            cout<<"Yes"<<endl;
        }
        else cout<<"No"<<endl;
    }
    else
    {
        if(t<=k)
        {
            cout<<"Yes"<<endl;
        }
        else cout<<"No"<<endl;
    }

    return 0;
}