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 (Clang 3.8.0) (judge verdict: Accepted)

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <string>
#include <sstream>
#include <complex>
#include <vector>
#include <list>
#include <queue>
#include <deque>
#include <stack>
#include <map>
#include <set>
#include <climits>

using namespace std;
typedef long long unsigned int ll;
typedef pair<int, int> P;

#define EPS (1e-7)
#define INF (1e12)
#define PI (acos(-1))

#define amari(x) ((x) % 1000000007)

#define fillInt(xs, x)        \
  for (int i = 0; i < (x); i++) \
    scanf("%d", &xs[i]);
#define fillLong(xs, x)       \
  for (int i = 0; i < (x); i++) \
    scanf("%ld", &xs[i]);
#define fillString(xs, x)       \
  for (int i = 0; i < (x); i++) \
    cin >> xs[i];
#define sortv(xs) sort(xs.begin(), xs.end());
#define sortvdi(xs) sort(xs.begin(), xs.end(), std::greater<int>());
#define lbv(xs, x) lower_bound(xs.begin(), xs.end(), x) - xs.begin();
#define ubv(xs, x) upper_bound(xs.begin(), xs.end(), x) - xs.begin();

#define rep(i,n) for(int i=0; i<(n); i++)
#define isValidPoint(x, y, mx, my) x >= 0 && x  < mx && y >= 0 && y < my

int dx[4] = {1, 0, -1, 0}, dy[4] = {0, 1, 0, -1};

int main()
{
  cin.tie(0);
  ios::sync_with_stdio(false);

  int n;

  cin >> n;

  long a[n];

  fillLong(a, n);

  int kcnt = 0;
  rep(i, n) {
    if (a[i] % 2) kcnt++;
  }
  
  cout << (kcnt % 2 ? "NO" : "YES") << endl;

  return 0;
}