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() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  vector<long long int> A(n);
  for (int i = 0; i < n; i++) cin >> A[i];
  int i = 0, j = n - 1;
  while (i < j) {
    if (!(A[i] & 1) && (A[j] & 1))
      i++, j--;
    else if ((A[i] & 1) && !(A[j] & 1))
      swap(A[i], A[j]), i++, j--;
    else if (!(A[i] & 1) && !(A[j] & 1))
      i++, swap(A[i], A[j]);
    else if ((A[i] & 1) && (A[j] & 1))
      j--, swap(A[i], A[j]);
  }
  int even_it;
  if (!(A[0] & 1)) {
    for (even_it = 0; !(A[even_it] & 1) && even_it < n; even_it++)
      ;
    sort(A.begin(), A.begin() + even_it);
    sort(A.begin() + even_it, A.end(), greater<long long int>());
  } else {
    even_it = -1;
    sort(A.begin(), A.end(), greater<long long int>());
  }
  int even_no, odd_no, t;
  if (even_it == -1)
    even_no = 0, odd_no = n;
  else
    even_no = even_it, odd_no = n - even_it;
  if ((even_no == odd_no) || abs(even_no - odd_no) == 1)
    cout << "0\n";
  else if (even_no > odd_no) {
    t = even_no - odd_no - 1, i = 0;
    long long int res = accumulate(A.begin(), A.begin() + t, 0);
    cout << res << "\n";
  } else {
    t = odd_no - even_no - 1, i = n - 1;
    long long int res = 0;
    while (t--) {
      res += A[i];
      i--;
    }
    cout << res << "\n";
  }
  return 0;
}
