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 a[2005], b[2005], c[2005];
int main() {
  int n, i, d = 0, j = 0, bsum = 0, csum = 0;
  cin >> n;
  for (i = 0; i < n; i++) {
    cin >> a[i];
    if (a[i] % 2 == 0) {
      b[d] = a[i];
      d++;
    } else {
      c[j] = a[i];
      j++;
    }
  }
  sort(c, c + j);
  sort(b, b + d);
  if (fabs(d - j) == 1 || d == j)
    cout << "0" << endl;
  else {
    if (d > j) {
      for (i = 0; i < d - j - 1; i++) {
        bsum += b[i];
      }
      cout << bsum;
      cout << endl;
    } else {
      for (i = 0; i < j - d - 1; i++) csum += c[i];
      cout << csum;
      cout << endl;
    }
  }
  return 0;
}
