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() {
  long long int a, b, c, d, e, f, g;
  vector<int> v1, v2;
  while (cin >> a) {
    for (b = 1; b <= a; b++) {
      cin >> c;
      v1.push_back(c);
    }
    v2 = v1;
    sort(v1.begin(), v1.end());
    reverse(v1.begin(), v1.end());
    for (b = 1; b <= a; b++) {
      d = 0;
      for (c = 0; c < v1.size(); c++) {
        if (v1[c] != 0 && v1[c] % 2 != 0 && b % 2 != 0) {
          v1[c] = 0;
          d = 1;
          break;
        } else if (v1[c] != 0 && v1[c] % 2 == 0 && b % 2 == 0) {
          v1[c] = 0;
          d = 1;
          break;
        }
      }
      if (d == 0) {
        break;
      }
    }
    e = 0;
    for (c = 0; c < v1.size(); c++) {
      e += v1[c];
    }
    for (b = 1; b <= a; b++) {
      d = 0;
      for (c = 0; c < v1.size(); c++) {
        if (v1[c] != 0 && v1[c] % 2 != 0 && b % 2 == 0) {
          v1[c] = 0;
          d = 1;
          break;
        } else if (v1[c] != 0 && v1[c] % 2 == 0 && b % 2 != 0) {
          v1[c] = 0;
          d = 1;
          break;
        }
      }
      if (d == 0) {
        break;
      }
    }
    f = 0;
    for (c = 0; c < v1.size(); c++) {
      f += v1[c];
    }
    cout << min(e, f) << endl;
    v1.clear();
    v2.clear();
  }
  return 0;
}
