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;
int main() {
  int n;
  cin >> n;
  int ip;
  vector<int> e, o;
  long long ans = 0;
  for (int i = 0; i < n; i++) {
    cin >> ip;
    ans += ip;
    if (ip % 2 == 0) {
      e.push_back(ip);
    } else {
      o.push_back(ip);
    }
  }
  sort(e.rbegin(), e.rend());
  sort(o.rbegin(), o.rend());
  int l1 = e.size(), l2 = o.size();
  if (l1 == 0)
    ans -= o[0];
  else if (l2 == 0)
    ans -= e[0];
  else {
    int turn;
    int i1 = 0, i2 = 0;
    if (e[0] > o[0])
      turn = 0;
    else
      turn = 1;
    for (int i = 1; i <= 2 * min(l1, l2); i++) {
      if (turn % 2 == 0)
        ans -= e[i1++];
      else
        ans -= o[i2++];
      turn++;
    }
    if (l1 < l2) {
      ans -= o[i2++];
    } else if (l1 > l2) {
      ans -= e[i1++];
    }
  }
  cout << ans << "\n";
  return 0;
}
