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() {
  int n;
  cin >> n;
  int a[2005];
  int odd[2005], even[2005];
  int cnt1 = 0, cnt2 = 0;
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    if (a[i] & 1) {
      cnt1++;
      odd[cnt1] = a[i];
    } else {
      cnt2++;
      even[cnt2] = a[i];
    }
    ans += a[i];
  }
  sort(odd + 1, odd + 1 + cnt1, greater<int>());
  sort(even + 1, even + 1 + cnt2, greater<int>());
  long long m = 0;
  for (int i = 1;; i++) {
    if (i & 1) {
      if (i / 2 > cnt1) break;
      m += odd[i / 2];
    } else {
      if (i / 2 > cnt2) break;
      m += even[i / 2];
    }
  }
  long long mm = 0;
  for (int i = 1;; i++) {
    if (i & 1) {
      if (i / 2 > cnt2) break;
      mm += even[i / 2];
    } else {
      if (i / 2 > cnt1) break;
      mm += odd[i / 2];
    }
  }
  ans = min(ans - m, ans - mm);
  printf("%lld", ans);
  return 0;
}
