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: Wrong Answer)

#include <bits/stdc++.h>

#define REP(i,n) for(int i=0;i<n;i++)
#define FOR(i,m,n) for(int i=m;i<n;i++)
#define FORR(i,m,n) for(int i=m;i>=n;i--)
#define pb(x) push_back(x) 
#define SORT(x) sort((x).begin(),(x).end())
#define INF 999999999999

using namespace std;

#define int long long

int32_t main(){
	int n;
	vector<int> a;
	cin >> n;
	int cnt_1=0, cnt_2;
	REP(i, n){
		int x;
		cin >> x;
		a.pb(x%2);
		
		if(x%2) cnt_1++;
	}
	cnt_2=n-cnt_1;
	
	if(cnt_1%2){
		cout << "NO" << endl;
		return 0;
	}
	else{
		cnt_1/=2;
		if(cnt_1%2){
			if(cnt_2%2){
				cout << "YES" << endl;	
				return 0;
			}
		}
		else{
			if(!(cnt_2%2)){
				cout << "YES" << endl;
				return 0;
			}
		}
	}
	
	cout << "NO" << endl;
}