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 long n, i, j, k;
  cin >> n;
  vector<long long> a(n);
  long long eve = 0, odd = 0;
  long long sum = 0;
  for (i = 0; i < n; i++) {
    cin >> a[i];
    if (a[i] % 2 == 0) {
      eve++;
    } else {
      odd++;
    }
    sum += a[i];
  }
  sort(a.begin(), a.end());
  if (eve == odd - 1 || odd == eve - 1 || odd == eve) {
    cout << 0;
    return 0;
  }
  if (eve == n || odd == n) {
    cout << sum - a[n - 1];
    return 0;
  }
  long long sum1 = 0;
  if (eve > odd) {
    long long xx = n - (odd + 1 + odd);
    long long c = 0;
    for (j = 0; j < n; j++) {
      if (c == xx) {
        break;
      }
      if (a[j] % 2 == 0) {
        sum1 += a[j];
        c++;
      }
    }
  } else {
    long long xx = n - (eve + eve + 1);
    long long c = 0;
    for (j = 0; j < n; j++) {
      if (c == xx) {
        break;
      }
      if (a[j] % 2 != 0) {
        sum1 += a[j];
      }
      c++;
    }
  }
  cout << sum1;
}
