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 OCaml (4.02.3) (judge verdict: Accepted)

open Scanf
open Printf
open String;;

(* sscanf (read_line ()) "%d %d" 
   (fun x y -> printf "%d" ((x-1)*(y-1)));; *)



(* let abb s = (sub s 0 1) 
            ^ (string_of_int (length s - 2))
            ^ (sub s (length s - 1) 1);;

sscanf (read_line ()) "%s"
          (fun s -> printf "%s" (abb s));; *)

let judge l =
  let len = List.length l in
  let f l = List.fold_left 
              (fun (x,l') y -> if y mod 4==0 then (x+1,l') else (x,y::l')) (0,[]) l in
  let g l = List.fold_left (fun x y -> x + if y mod 2!=0 then 1 else 0) 0 l in
  let (four, fourl) = f l in
  if four >= len/2 then true else
    if g fourl <= four then true else false

let usage n = 
  let list = ref [] in
  for i = 1 to n do
    list := (scanf (if i < n then "%d " else "%d") (fun x -> x))::!list
  done; !list;;

sscanf (read_line ()) "%d" 
   (fun x -> printf "%s" (if judge (usage x) then "Yes" else "No"));;