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>
#pragma GCC optimize("O3")
using namespace std;
using ll = long long;
using ld = long double;
using vi = vector<int>;
using vll = vector<ll>;
using pll = pair<ll, ll>;
using pii = pair<int, int>;
const double pi = acos(-1.0);
int main() {
  int n, odd = 0, evn = 0;
  cin >> n;
  vector<int> a(n, 0);
  vector<int> e, o;
  for (int i = 0; i < (n); i++) {
    cin >> a[i];
    if (a[i] % 2 == 0) {
      evn++;
      e.push_back(a[i]);
    } else {
      odd++;
      o.push_back(a[i]);
    }
  }
  sort(a.begin(), a.end());
  sort(e.begin(), e.end());
  sort(o.begin(), o.end());
  reverse(a.begin(), a.end());
  if (evn > odd) {
    int s = 0, d = evn - odd;
    if (d == 1) {
      cout << 0;
      return 0;
    }
    for (int i = 0; i < d - 1; i++) {
      s += e[i];
    }
    cout << s;
  } else {
    int s = 0, d = odd - evn;
    if (d == 1) {
      cout << 0;
      return 0;
    }
    for (int i = 0; i < d - 1; i++) {
      s += o[i];
    }
    cout << s;
  }
  return 0;
}
