Problem:
Score : 300 points

Problem Statement

There are N integers written on a blackboard. The i-th integer is A_i.

Takahashi will repeatedly perform the following operation on these numbers:

Select a pair of integers, A_i and A_j, that have the same parity (that is, both are even or both are odd) and erase them.

Then, write a new integer on the blackboard that is equal to the sum of those integers, A_i+A_j.

Determine whether it is possible to have only one integer on the blackboard.

Constraints

2 ≦ N ≦ 10^5

1 ≦ A_i ≦ 10^9

A_i is an integer.

Input

The input is given from Standard Input in the following format:

N
A_1 A_2 … A_N

Output

If it is possible to have only one integer on the blackboard, print YES. Otherwise, print NO.

Sample Input 1

3
1 2 3

Sample Output 1

YES

It is possible to have only one integer on the blackboard, as follows:

Erase 1 and 3 from the blackboard, then write 4. Now, there are two integers on the blackboard: 2 and 4.

Erase 2 and 4 from the blackboard, then write 6. Now, there is only one integer on the blackboard: 6.

Sample Input 2

5
1 2 3 4 5

Sample Output 2

NO

Solution in OCaml (4.02.3) (judge verdict: Accepted)

let id x = x
let id' x = 1
let n = Scanf.scanf "%d\n" id

let g = (fun () -> Scanf.scanf "%d\n" id)
  
let repeat f =
  let rec loop acc = function
    | 0 -> List.rev acc
    | n -> loop (f () :: acc) (n - 1) in
  loop []
 
let r = repeat (fun () -> Scanf.scanf "%d " id) n
  
let rec check = function
  | [a] -> true
  | a :: rest ->
     match rest with
     | [b] -> if (a mod 2) = (b mod 2) then true else false
     | b :: rest2 ->
	match rest2 with
	| c :: rest3 -> if (a mod 2) = (b mod 2) then check ((a+b) :: rest2) else
	    if (a mod 2) = (c mod 2) then check ((a+c) :: b :: rest3) else check ((b+c) :: a :: rest3)

let ansbool = check r

let ans = if ansbool = true then "YES" else "NO"
    
let () = Printf.printf "%s\n" ans