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>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#define rep(i,n) for (long long i=0; i < (n); ++i)
#define rep2(i,n,m) for(long long i=n;i<=m;i++)
#define ALL(a)  (a).begin(),(a).end()
using namespace std;
using ll = long long;
using P = pair<int,int>;
const ll INF=1e18 ;
inline void chmax(ll& a,ll b){a=max(a,b);}
inline void chmin(ll& a,ll b){a=min(a,b);}
int main() {
 ll n ;
 cin >> n ;
 vector<ll> A(n) ;
 rep(i,n) cin >>A[i] ;
 ll ans=0 ;
 rep(i,n) if(A[i]%2==1) ans++ ;
  if(ans%2==0) cout<< "Yes"<<endl ;
  else cout <<"No"<<endl ;
      return 0;
}