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 Kotlin (1.0.0) (judge verdict: Accepted)

import java.util.*
 
fun main(args: Array<String>) {
    val sc = Scanner(System.`in`)
    val n = sc.nextInt()
    val a = (0 until n).map { sc.next().toLong() }
    println(problem069c(n, a))
}
 
fun problem069c(n: Int, a: List<Long>): String {
    val fourCount = a.count { it % 4 == 0L }
    val twoCount = a.count { it % 4 == 2L }
    val oneCount = a.count { it % 2 == 1L }
    // 「2の倍数だけど、4の倍数ではない」数はたくさんあっても1つあるのとおなじ
    // if(twoCount >= 1) 1 else 0
  
    // このままだと4の倍数にならない数たちの個数
    // oneCount + ( if(twoCount >= 1) 1 else 0 )
  
    // これらを4の倍数のとなりに置く
    // 4の倍数のとなりは fourCount + 1 個
  
    // 例えば...
    // [1] 4 [5] 4 [2 2 2 6 10] 8 [7] 12 [9]
    return if (oneCount + ( if(twoCount >= 1) 1 else 0 ) <= fourCount + 1) "Yes" else "No"
}