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 Rust (1.42.0) (judge verdict: Wrong Answer)

use proconio::{
    fastout, input,
    marker::{Bytes, Chars, Isize1, Usize1},
};

fn main() {
    input! {
        n:usize,
        a:[usize;n]
    }
    let mut even = 0_usize;
    let mut odd = 0;
    for i in a {
        if i % 2 == 0 {
            even += 1;
        } else {
            odd += 1;
        }
    }

    if even + odd == 1 {
        println!("YES");
        return;
    }

    if odd % 2 != 0 {
        println!("NO");
        return;
    }
    even += odd / 2;
    if even.count_ones() != 1 {
        println!("NO");
        return;
    }
    println!("YES");
}