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)

/*
　　こんにちは。あたしはｶｳｶﾞｰﾙ。
　　redcoderになるためAtCoderを巡る旅をしてます。

　　　 　 ＿_
　　　 ヽ|＿_|ノ　　　　ﾓｫ
　　　　||‘‐‘||ﾚ　　_)_, ―‐ 、
　　　　/(Ｙ (ヽ＿ /・ ヽ　　 ￣ヽ
　　　　∠ ＿ ゝ　 ｀^ヽ ﾉ.::::_(ノヽ
　　　　 _/ヽ　 　　  /ヽ￣￣/ヽ
*/

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define dump(x)  cout << #x << " = " << (x) << endl
#define YES(n) cout << ((n) ? "YES" : "NO"  ) << endl
#define Yes(n) cout << ((n) ? "Yes" : "No"  ) << endl
#define POSSIBLE(n) cout << ((n) ? "POSSIBLE" : "IMPOSSIBLE"  ) << endl
#define Possible(n) cout << ((n) ? "Possible" : "Impossible"  ) << endl

#define rep(i, n) REP(i, 0, n)                              // 0, 1, ..., n-1
#define REP(i, x, n) for(int i = x; i < n; i++)             // x, x + 1, ..., n-1
#define FOREACH(x,a) for(auto& (x) : (a) )

#define ALL(v) (v).begin() , (v).end()
#define RALL(v) (v).rbegin(), (v).rend()

#define COUT(x) cout << (x) << endl

template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }

int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);

    int n; cin >> n;
    int odd = 0, even = 0;
    rep(i, n){
        int a; cin >> a;
        if(a & 1) odd++;
        else even++;
    }
    if(odd == 1) {
        cout << "NO" << endl;
        return 0;
    }
    if((odd / 2 + even) & 1) cout << "NO" << endl;
    else cout << "YES" << endl;
}