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;
#define INF 2000000000
typedef long long int ll;
typedef pair<ll,ll> ii;
typedef pair<ii,ll> iii;
typedef vector<ll> vi;
typedef vector<ii> vii;
#define getbit(n,i) (((n)&(1LL<<(i)))!=0)
#define setbit0(n,i) ((n)&(~(1LL<<(i))))
#define setbit1(n,i) ((n)|(1LL<<(i)))
#define lastone(n) ((n)&(-(n)))
#define read freopen("debug\\in.txt","r",stdin)
#define write freopen("debug\\out.txt","w",stdout)
#define DBG(a) cout<<#a<<" ->->->-> "<<a<<"\n"
#define EN cout<<"\n"
#define fi first
#define se second
#define PI (acos(-1))
#define fastread ios::sync_with_stdio(false);cin.tie(NULL)
#define mod (1000000007)
#define asz 200005
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
template<class T,class V> ostream& operator<<(ostream &s,pair<T,V> a)
{
    s<<a.fi<<' '<<a.se;
    return s;
}
int main()
{
    fastread;
    ll T=1;
//    cin>>T;
    for(ll qq=1;qq<=T; qq++)
    {
        ll n;
        cin>>n;
        ll od=0;
        while(n--){
            ll x;
            cin>>x;
            od+=x%2;
        }
        if(od%2==0)cout<<"YES"<<endl;
        else cout<<"NO"<<endl;
    }
}