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++14 (GCC 5.4.1) (judge verdict: Accepted)

#pragma GCC optimize("Ofast")
#include <bits/stdc++.h>
#define rep(i,n) for (int i = 0; i < (n); ++i)
#define REP(i,a,b) for (int i = a; i < (b); ++i)
#define all(x) (x).begin(),(x).end()
const int INF = 1000000007;
typedef long long ll;
using namespace std;
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }
ll gcd(ll a, ll b) { return b ? gcd(b, a%b) : a; }
int main(){
  // ios::sync_with_stdio(false);cin.tie(nullptr);
  int n;
  scanf("%d",&n);
  int tmp;
  int count = 0;
  rep(i,n){
    scanf("%d",&tmp);
    if(tmp % 2 != 0)count++;
  }
  if(count % 2 == 0)printf("YES\n");
  else printf("NO\n");
  return 0;
}