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 <algorithm>
#include <queue>
#include <vector>
#include <set>
#include <cmath>
#include <string>
#define FOR(i,a,b) for(int i=a;i<b;i++)
#define ROF(i,a,b) for(int i=b-1;i>=a;i--)
#define FI first
#define SE second
#define SI size()
#define MA(i,j) make_pair(i,j)
#define PA pair<int,int>
#define PF push_front
#define PB push_back
#define VE vector<int>
#define VP vector<PA>
#define LL long long
#define YES(i) cout<<(i?"YES":"NO")<<endl
#define Yes(i) cout<<(i?"Yes":"No")<<endl
using namespace std;
//
const int INF=1e9+7;
const int MOD=1e9+7;
//
int main(){
  int N;
  int A;
  cin>>N;
  int odd=0,even=0;
  FOR(i,0,N){
    cin>>A;
    if(A%2){
      odd++;
    }else{
      even++;
    }
  }
  if(odd%2==0&&even%2==0){
    YES(1);
  }else if(odd%2==0&&even%2==1){
    YES(1);
  }else if(odd%2==1&&even%2==0){
    YES(0);
  }else if(odd%2==1&&even%2==1){
    YES(0);
  }
  return 0;
}