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;
const int N = 2e5 + 5;
long long ara[2005], n, sum, jur, bejur;
vector<int> even, odd;
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> ara[i];
    if (ara[i] & 1) {
      odd.push_back(ara[i]);
      bejur++;
    } else {
      even.push_back(ara[i]);
      jur++;
    }
    sum += ara[i];
  }
  if (jur == bejur) {
    cout << "0\n";
    exit(0);
  }
  if (abs(jur - bejur) == 1) {
    cout << "0\n";
    exit(0);
  }
  long long mn = min(jur, bejur);
  if (odd.size() > 0) sort(odd.rbegin(), odd.rend());
  if (even.size() > 0) sort(even.rbegin(), even.rend());
  if (mn > 0) {
    for (int i = 0; i < mn; i++) {
      sum -= odd[i];
      sum -= even[i];
    }
    if (odd[mn] > even[mn]) {
      sum -= odd[mn];
    } else {
      sum -= even[mn];
    }
  } else {
    if (jur > bejur) {
      sum -= even[0];
    } else {
      sum -= odd[0];
    }
  }
  cout << sum << "\n";
  return 0;
}
