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 <iostream>
#include <vector>
#include <string>
#include <array>
#include <queue>
#include <deque>
#include <set>
#include <list>
#include <map>
#include <stack>
#include <utility>
#include <algorithm>
#include <numeric>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cctype>
#include <cstring>
#include <climits>
#include <bitset>
#include <random>
#include <functional>
#include <sstream>
#include <iomanip>

using namespace std;

#define _CRT_SECURE_NO_WARNINGS
#define REP(i, n) for(LL i=0; i<(n); i++)
#define FOR(i, m, n) for(LL i=(m);i<(n);i++)
#define SZ(x) ((int)(x).size())
#define SORT(x) sort((x).begin(),(x).end())
#define REVE(x) reverse((x).begin(),(x).end())
#define MP make_pair
#define PB push_back
#define INF INT_MAX

typedef vector<int> VI;
typedef vector<string> VS;
typedef vector<vector<int>> VVI;
typedef pair<int, int> PII;
typedef long long LL;

int main() {
	int n, e = 0, o = 0;
	cin >> n;
	REP(i, n) {
		int x; cin >> x;
		if (x % 2 == 0)e++;
		else o++;
	}

	bool flag;
	if (o % 2 == 1)flag = false;
	else flag = true;
	puts(flag ? "YES" : "NO");
}