Polycarp has an array a consisting of n integers.

He wants to play a game with this array. The game consists of several moves. On the first move he chooses any element and deletes it (after the first move the array contains n-1 elements). For each of the next moves he chooses any element with the only restriction: its parity should differ from the parity of the element deleted on the previous move. In other words, he alternates parities (even-odd-even-odd-... or odd-even-odd-even-...) of the removed elements. Polycarp stops if he can't make a move.

Formally: 

  * If it is the first move, he chooses any element and deletes it; 
  * If it is the second or any next move: 
    * if the last deleted element was odd, Polycarp chooses any even element and deletes it; 
    * if the last deleted element was even, Polycarp chooses any odd element and deletes it. 
  * If after some move Polycarp cannot make a move, the game ends. 



Polycarp's goal is to minimize the sum of non-deleted elements of the array after end of the game. If Polycarp can delete the whole array, then the sum of non-deleted elements is zero.

Help Polycarp find this value.

Input

The first line of the input contains one integer n (1 ≤ n ≤ 2000) — the number of elements of a.

The second line of the input contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 10^6), where a_i is the i-th element of a.

Output

Print one integer — the minimum possible sum of non-deleted elements of the array after end of the game.

Examples

Input


5
1 5 7 8 2


Output


0


Input


6
5 1 2 4 6 3


Output


0


Input


2
1000000 1000000


Output


1000000
#include <bits/stdc++.h>
using namespace std;
const long long N = 2020;
long long Try(vector<long long>& v1, vector<long long>& v2);
long long a[N];
vector<long long> v[2];
int main() {
  long long n;
  cin >> n;
  for (long long i = 1; i <= n; i++)
    scanf("%I64d", a + i), v[a[i] & 1].push_back(a[i]);
  for (long long k : {0, 1}) sort(v[k].begin(), v[k].end());
  cout << min(Try(v[0], v[1]), Try(v[1], v[0]));
}
long long Try(vector<long long>& v1, vector<long long>& v2) {
  long long Turn = 1;
  while (true) {
    vector<long long>& v3 = Turn ? v1 : v2;
    if (v3.empty()) break;
    v3.pop_back();
    Turn = !Turn;
  }
  long long Sum = 0;
  for (auto x : v1) Sum += x;
  for (auto x : v2) Sum += x;
  return Sum;
}
