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;
  cin >> n;
  long int var;
  vector<long int> even, odd, even1, odd1;
  while (n--) {
    cin >> var;
    if (var % 2 == 0)
      even.push_back(var);
    else
      odd.push_back(var);
  }
  sort(even.begin(), even.end());
  sort(odd.begin(), odd.end());
  even1 = even;
  odd1 = odd;
  while (1) {
    if (even.empty())
      break;
    else
      even.pop_back();
    if (odd.empty())
      break;
    else
      odd.pop_back();
  }
  while (1) {
    if (odd1.empty())
      break;
    else
      odd1.pop_back();
    if (even1.empty())
      break;
    else
      even1.pop_back();
  }
  long int ans1, ans2;
  ans1 = ans2 = 0;
  for (long int i = 0; i < even.size(); i++) ans1 += even[i];
  for (long int i = 0; i < odd.size(); i++) ans1 += odd[i];
  for (long int i = 0; i < even1.size(); i++) ans2 += even1[i];
  for (long int i = 0; i < odd1.size(); i++) ans2 += odd1[i];
  cout << min(ans1, ans2);
  return 0;
}
