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: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;

#define rp(i,n) for(int i=0;i<n;i++)
#define rep(i,m,n) for(int i=m;i<=n;i++)
#define llrp(i,n) for(long long i=0;i<n;i++)
#define llrep(i,m,n) for(long long i=m;i<=n;i++)
#define sort(a) sort(a.begin(),a.end())
#define reverse(a) reverse(a.begin(),a.end())
#define pb push_back
#define elif else if
#define vcin(a); rp(i,a.size()){cin>>a[i];}
#define vcout(a); rp(i,a.size()){cout<<a[i]<<endl;}
typedef vector<int> vi;
typedef vector<long long> vll;
typedef vector<string> vs;
typedef vector<char> vc;
typedef long long ll;
typedef string S;
typedef pair<int,int> P;

bool isOK(int index, int key,vi a) {
    if (a[index] >= key) return true;
    else return false;
}

int binary_search(int key,vi a) {
    int left = -1;
    int right = (int)a.size();
    while (right - left > 1) {
        int mid = left + (right - left) / 2;
        
        if (isOK(mid, key,a)) right = mid;
        else left = mid;
    }
    return right;
}

int modpow(int a,int b,int p){
    if(b==0){
        return 1;
    }elif(b%2==0){
        return modpow((a*a)%p,b/2,p);
    }else{
        return (a*modpow(a,b-1,p))%p;
    }
}

int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    int n,e=0,o=0;
    cin >> n;
    vi a(n);
    rp(i,n){
        cin >> a[i];
        a[i]%=2;
        if(a[i]){
            o++;
        }else{
            e++;
        }
    }
    cout << (o%2==0&&(e+o/2)%2==0?"YES":"NO") << endl;
    return 0;
}