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;
using ll = long long;

void init();
void solve();

#define rep(i, a) for (int i = 0; i < (int) (a); i++)
#define reps(i, a, b) for (int i = (int) (a); i < (int) (b); i++)
#define rrep(i, a) for (int i = (int) a-1; i >= 0; i--)
#define rreps(i, a, b) for (int i = (int) (a)-1; i >= (int) (b); i--)
#define MP(a, b) make_pair((a), (b))
#define PB(a) push_back((a))
#define all(v) (v).begin(), (v).end()
#define PRINT(f) if((f)){cout << (TRUE__) << endl;}else{cout << FALSE__ << endl;}
#define TRUE__  "YES"
#define FALSE__ "NO"
#define PERM(v) next_permutation(all(v))
#define UNIQUE(v) sort(all(v));(v).erase(unique(all(v)), v.end())
#define CIN(type, x) type x;cin >> x

#ifdef LOCAL
#define lcout(a) cout << a
#define lcoutln(a) cout << a << endl
#define lcerr(a) cerr << a
#define lcerrln(a) cerr << a << endl
#else
#define lcout(a) 
#define lcoutln(a) 
#define lcerr(a) 
#define lcerrln(a) 
#endif
#define MOD 1000000007
#define PI 3.141592653589793238462643383

signed main()
{
	init();
	solve();
}

void init()
{
	int N;
	cin >> N;
	int res = 0;
	rep(i, N) {
		int a;
		cin >> a;
		res += a % 2;
	}
	PRINT(res % 2 == 0);
}

void solve()
{
}