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 <algorithm>
#include <cmath>
#include <iostream>
#include <iomanip>
#include <map>
#include <string>
#include <vector>
#include <queue>
#include <stack>
#include <set>
#include <list>
#define INF 1000000000
#define MOD 1000000007ll
#define EPS 1e-10
#define REP(i,m) for(long long i=0; i<m; i++)
#define FOR(i,n,m) for(long long i=n; i<m; i++)
#define DUMP(a) for(long long dump=0; dump<(ll)a.size(); dump++) { cout<<a[dump]; if(dump!=(ll)a.size()-1) cout<<" "; else cout<<endl; }
#define ALL(v) v.begin(),v.end()
#define pb push_back
using namespace std;
typedef long long int ll;
typedef pair<ll, ll> P;
typedef long double ld;
using namespace std;

int main(){
	int n;
	cin >> n;
	int ans = 0;
	REP(i, n){
		int x;
		cin >> x;
		if(x % 2 == 0) ans++;
	}
	cout << ans << endl;
	return 0;
}