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++ (GCC 9.2.1) (judge verdict: Accepted)

/*---------------Go Code GO---------------*/

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

using namespace std;
using namespace __gnu_pbds;

template <typename T>
using Order_Set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
//(order) Set.order_of_key(); (pointer) Set.find_by_order();

#define PI acos(-1.0)
#define O_O                           \
    ios_base::sync_with_stdio(false); \
    cin.tie(NULL)
#define precision(a) fixed << setprecision(a)
#define endl '\n'
#define Flush cout << flush
#define LLMX 0x3fffffffffffffff

long long N;

int main()
{
    O_O;
    long long i, j, k, l, tst = 0, ans = 0, csno = 0;
    long long odd = 0, even = 0;

    cin >> N;

    for (int i = 0; i < N; i++)
    {
        cin >> k;
        odd += (k % 2);
        even += (k % 2 == 0);
    }
    if (odd % 2 == 0)
        cout << "YES" << endl;
    else
        cout << "NO" << endl;

    return 0;
}