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 int n, i;
  cin >> n;
  long int a[n];
  for (i = 0; i < n; i++) cin >> a[i];
  long int o = 0;
  long int e = 0;
  for (i = 0; i < n; i++) {
    if (a[i] % 2 == 0)
      e++;
    else
      o++;
  }
  sort(a, a + n);
  long int k;
  if ((o == e) || (max(o, e) - min(o, e) == 1))
    cout << "0";
  else if (o == 0 || e == 0) {
    long int sum = 0;
    for (i = 0; i < n - 1; i++) {
      sum = sum + a[i];
    }
    cout << sum;
  } else {
    if (o > e) {
      long int sum = 0;
      k = o - e - 1;
      for (i = 0; k > 0; i++) {
        if (a[i] % 2 != 0) {
          sum = sum + a[i];
          k--;
        }
      }
      cout << sum;
    } else {
      long int sum = 0;
      k = e - o - 1;
      for (i = 0; k > 0; i++) {
        if (a[i] % 2 == 0) {
          sum = sum + a[i];
          k--;
        }
      }
      cout << sum;
    }
  }
}
