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)

#include<bits/stdc++.h>

using namespace std;

#define pb push_back
#define mp make_pair
#define F first
#define S second

#define fo(i, n) for(int i = 1; i <= n; ++i)
#define rep(i, a, b) for(int (i) = (a); (i) < (b); ++(i))

typedef long long ll;
typedef pair <int, int> pii;
typedef pair <ll, ll> pll;

const int N = 200200;
const int mod = 1e9 + 7;
const ll INF = 1e18;
int n,cnt;

int main()
{
	ios::sync_with_stdio(0); cin.tie(0);
		cin>>n;
		fo(i,n)
		{
			int x;
			cin>>x;
			if(x%2==1)
				++cnt;
		}
		if(cnt&1)
		cout<<"NO";
		else cout<<"YES";
	return 0;	
}