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<bits/stdc++.h>
using namespace std;
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#pragma GCC optimize ("-ffloat-store")  
#pragma GCC optimize ("-fno-defer-pop")
#define all(a) a.begin(),a.end()
#define bs binary_search
#define ll long long int
#define ld long double
ll power(ll a,ll b,ll m){ if(b==0) return 1; if(b==1) return a%m; ll t=power(a,b/2,m)%m; t=(t*t)%m; if(b&1) t=((t%m)*(a%m))%m; return t;}
ll modInverse(ll a, ll m) { return power(a, m-2, m); }
#define ipr pair<ll,ll>
#define ps push_back
#define fs first
#define sc second
#define mkp make_pair
#define mod 1000000007
int main(){
	ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);
	
	ll i,j,k,n;
	cin>>n;
	ll ar[n];
	for(i=0;i<n;i++){
		cin>>ar[i];
	}
	k=0;
	for(i=0;i<n;i++){
		if(ar[i]%2==1){
			k++;
		}
	}
	if(k%2==0){
		cout<<"YES";
	}
	else {
		cout<<"NO";
	}
	return 0;
}