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 <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <stack>
#include <queue>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cctype>
#include <string>
#include <cstring>
#include <ctime>
 
using namespace std;
 
typedef long long ll;
typedef unsigned long long ull;
 
static const double EPS = 1e-8;
static const double PI = 4.0 * atan(1.0);
static const double PI2 = 8.0 * atan(1.0);
 
#define REP(i,n)	for(int i=0;i<(int)n;++i)
#define ALL(c)		(c).begin(),(c).end()
#define CLEAR(v)	memset(v,0,sizeof(v))
#define MP(a,b)		make_pair((a),(b))
#define ABS(a)		((a)>0?(a):-(a))
#define FOR(i,s,n)	for(int i=s;i<(int)n;++i)

int main(){
    ll n;
    cin >> n;
    vector<ll> a(n);
    for(int i=0;i<n;i++) cin >> a[i];
    ll cnt = 0;
    for(int i=0;i<n;i++) {
        if(a[i]%2==1)
            cnt++;
    }
    if(cnt%2==1)
        cout << "NO" << endl;
    else
        cout << "YES" << endl;

}