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>
#include <ext/hash_map>
#include <ext/numeric>

using namespace std;
using namespace __gnu_cxx;

#define REP(i,n) for( (i)=0 ; (i)<(n) ; (i)++ )
#define rep(i,x,n) for( (i)=(x) ; (i)<(n) ; (i)++ )
#define REV(i,n) for( (i)=(n) ; (i)>=0 ; (i)-- )
#define FORIT(it,x) for( (it)=(x).begin() ; (it)!=(x).end() ; (it)++ )
#define foreach(it,c) for(__typeof((c).begin()) it=(c).begin();it!=(c).end();++it)
#define rforeach(it,c) for(__typeof((c).rbegin()) it=(c).rbegin();it!=(c).rend();++it)
#define foreach2d(i, j, v) foreach(i,v) foreach(j,*i)
#define all(x) (x).begin(),(x).end()
#define rall(x) (x).rbegin(),(x).rend()
#define SZ(x) ((int)(x).size())
#define MMS(x,n) memset(x,n,sizeof(x))
#define mms(x,n,s) memset(x,n,sizeof(x)*s)
#define pb push_back
#define mp make_pair
#define NX next_permutation
#define UN(x) sort(all(x)),x.erase(unique(all(x)),x.end())
#define CV(x,n) count(all(x),(n))
#define FIND(x,n) find(all(x),(n))-(x).begin()
#define ACC(x) accumulate(all(x),0)
#define PPC(x) __builtin_popcountll(x)
#define LZ(x) __builtin_clz(x)
#define TZ(x) __builtin_ctz(x)
#define mxe(x) *max_element(all(x))
#define mne(x) *min_element(all(x))
#define low(x,i) lower_bound(all(x),i)
#define upp(x,i) upper_bound(all(x),i)
#define NXPOW2(x) (1ll << ((int)log2(x)+1))
#define PR(x) cout << #x << " = " << (x) << endl ;

typedef unsigned long long ull;
typedef long long ll;
typedef vector<int> vi;
typedef vector<vector<int> > vvi;
typedef pair<int, int> pii;

const int OO = (int) 2e9;
const double eps = 1e-9;

int main() {
	std::ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	cout.tie(NULL);
#ifndef ONLINE_JUDGE
//	freopen("in.txt", "rt", stdin);
//	freopen("out.txt", "wt", stdout);
#endif
	int n, x, cnt = 0;
	cin >> n;
	for (int i = 0 ;i < n; i++) {
		cin >> x;
		if (x & 1)
			cnt++;
	}
	(cnt & 1) ? puts("NO") : puts("YES");
	return 0;
}