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 C (GCC 9.2.1) (judge verdict: Runtime Error)

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int main(void) {
    char buf[1000000000];
    int N;
    char* a;
    int Odd = 0;

    fgets(buf, sizeof(buf), stdin);
    sscanf(buf, "%d\n", &N);
    fgets(buf, sizeof(buf), stdin);
    a = strtok(buf, " ");
    if((atoi(a))%2)
      Odd++;
  
    while ((a = strtok(NULL, " ")) != NULL){
      if((atoi(a))%2)
      Odd++;
    }
  if(Odd%2)
    printf("NO");
  else
    printf("YES");
  
  printf("%d",Odd);
    return 0;
}