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>
 
#define pii pair<int, int>
#define f first
#define s second
#define pb push_back
#define mp make_pair
#define ll long long 
#define ull unsigned long long 
#define ld long double
#define sz(v) int(v.size())
#define all(v) v.begin(), v.end()
 
#define y1 what
 
using namespace std;
 
const int N = (int) 1e5 + 10;
const int M = (int) 101;
const ll big =  (1LL << 51);
const ll LINF = (ll) 1e18;
const int INF = (int) 1e9 + 7;
const int INF1 = (int) 1e9 + 3;
const double EPS = (double) 1e-6;
const double PI =  3.14159265359;
 
int n;
 
int main() {
    #define fn "balls"
    #ifdef witch
        freopen("input.txt", "r", stdin);
        freopen("output.txt", "w", stdout);
    #else
//        freopen(fn".in", "r", stdin);
//        freopen(fn".out", "w", stdout);
    #endif
    srand(time(0));
    cin >> n;
    int ans = 0;
    while (n--) {
        int x;
        cin >> x;
        ans ^= x % 2;
    }
    cout << (!ans ? "YES" : "NO");
    return 0;
}