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)

#include <bits/stdc++.h>

using namespace std;

typedef long long ll;

#define REP(i, a, b) for (i = a; i < b; i++)
#define REP_ITER(it, v) for (it = v.begin(); it != v.end(); it++)

#define FASTIO                   \
    ios::sync_with_stdio(false); \
    cin.tie(0)

#define TEST(test) \
    int test;      \
    cin >> test;   \
    while (test--)

#define cll(a, b) cin >> a >> b
#define YES cout << "YES\n"
#define NO cout << "NO\n"

#define vi vector<int>
#define vi_iterator vector<int>::iterator

#define si set<int>
#define si_iterator set<int>::iterator

double pi = 2 * acos(0.0);
int i, j;
int main()
{
    FASTIO;
    ll n,odd=0;
    cin >> n;
    ll arr[n];
    REP(i,0,n){
        cin >> arr[i];
        if(arr[i]%2==1){
            odd++;
        }
    }
    if(odd%2==1)NO;
    else YES;
}