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 <iostream>
#include <algorithm>
#include <vector>
#include <set>
#include <map>
#include <unordered_map>
#include <unordered_set>
#include <bitset>
#include <chrono>
#include <ctime>
#include <queue>
#include <math.h>
#include <deque>
#include <stack>
#include <iomanip>
#include <assert.h>
#include <stdio.h>
#include <cstring>
#include <random>

using namespace std;
#define int long long
#define ll long long
#define ld long double
#define ull unsigned long long
#define endl '\n'
#define try tr
#define pii pair<int, int>
#define ti tuple <int, int, int>
#define vi vector <int>
#define vpi vector <pii>
#define umap unordered_map
#define uset unordered_set
#define left left11
#define right right11
#define all(c) (c).begin(), (c).end()
#define pb push_back
#define sqr(a) ((a) * (a))
#define fast ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);

#define fs first
#define sc second

ld TL = 0.25;
bool is_tl = 0;
int CALLS = 0;

inline bool IS() {
    if (++CALLS == 1000) {
        CALLS = 0;
        is_tl |= clock() > (TL - 0.1) * CLOCKS_PER_SEC;
    }
    return is_tl;
}

template<typename T1, typename T2>
inline void amin(T1 &a, T2 b) { if (a > b) a = b; }

template<typename T1, typename T2>
inline void amax(T1 &a, T2 b) { if (a < b) a = b; }

template<typename T1, typename T2>
ostream &operator<<(ostream &os, pair<T1, T2> &p) {
    os << p.first << ' ' << p.second << endl;
    return os;
}

template<typename T1, typename T2>
istream &operator>>(istream &is, pair<T1, T2> &p) {
    is >> p.first >> p.second;
    return is;
}

template<typename T>
ostream &operator<<(ostream &os, vector<T> &v) {
    for (auto &u : v) os << u << ' ';
    os << endl;
    return os;
}

mt19937 rnd(chrono::steady_clock::now().time_since_epoch().count());
const ll N = 1e6 + 7, NS = 2e3 + 7, lg = 19, sq = 550, inf = 1e18 + 7, SM = 1e2 + 7;
const ld eps = 1e-9;
int hash_mod[4] = {1000000007, 998244353, 1000000009, 999999937}, mod = hash_mod[rnd() % 4];
int hash_pows[4] = {179, 239, 1007, 2003}, P = hash_pows[rnd() % 4];
void solve() {
    int n, c1 = 0, c0 = 0;
    cin >> n;
     vi a(n);
     for (int i= 0; i < n; i++) {
         cin >> a[i];
         if(a[i] % 2)
             c1++;
         else
             c0++;
     }
     if(c1 % 2 == 0)
         cout << "YES\n";
     else
         cout << "NO\n";
}

signed main() {
    fast
    solve();
}