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 (1.15.1) (judge verdict: Wrong Answer)

use std::io::{BufRead, Write, stdin, stdout};

fn solve<R: BufRead, W: Write>(mut input: R, mut output: W) {
    let mut s = String::new();
    input.read_line(&mut s).unwrap();
    let n: u64 = s.split_whitespace().next().unwrap().parse().unwrap();
    s.clear();
    let mut a: Vec<i64> = Vec::with_capacity(n as usize);
    input.read_line(&mut s).unwrap();
    for buf in s.split_whitespace() {
        a.push(buf.parse().unwrap());
    }
    let sum: i64 = a.iter().fold(0, |acc, &x| acc + x);
    let cnt: i64 = sum / ((1 + n) * n / 2) as i64;
    let mut result = (sum % ((1 + n) * n / 2) as i64) == 0;
    let first = a[0];
    for i in 1..a.len() {
        if (a[i] - i as i64 * cnt - first) % n as i64 != 0 {
            result = false;
            break;
        }
    }
    for i in a {
        if i < cnt || i > cnt * n as i64 {
            result = false;
            break;
        }
    }
    writeln!(output, "{}", if result { "YES" } else { "NO" }).unwrap();
}

fn main() {
    let stdin = stdin();
    solve(stdin.lock(), stdout());
}

#[cfg(test)]
mod tests {
    fn test(input: &str, output: &str) {
        let mut v = Vec::<u8>::new();
        ::solve(input.as_bytes(), &mut v);
        assert_eq!(String::from_utf8(v).unwrap(), output);
    }
    #[test]
    fn case1() {
        test("5
4 5 1 2 3\n",
             "YES\n");
    }
    #[test]
    fn case2() {
        test("5
6 9 12 10 8\n",
             "YES\n");
    }
    #[test]
    fn case3() {
        test("4
1 2 3 1\n",
             "NO\n");
    }
}