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>
long long int arr[2005], temp[2005];
void sort(int l, int r) {
  if (l >= r) return;
  int m = (l + r) / 2;
  int l1 = l, l2 = m + 1, r1 = m, r2 = r, i = l;
  sort(l1, r1);
  sort(l2, r2);
  while (l1 <= r1 || l2 <= r2) {
    if (l1 > r1) {
      temp[i++] = arr[l2++];
    } else if (l2 > r2) {
      temp[i++] = arr[l1++];
    } else if (arr[l1] <= arr[l2]) {
      temp[i++] = arr[l1++];
    } else {
      temp[i++] = arr[l2++];
    }
  }
  for (i = l; i <= r; i++) {
    arr[i] = temp[i];
  }
  return;
}
int main(void) {
  int N, i, cev = 0, cod = 0;
  scanf("%d", &N);
  for (i = 0; i < N; i++) {
    scanf("%lld", &arr[i]);
    if (arr[i] % 2 == 0)
      cev++;
    else
      cod++;
  }
  int f = 0, c;
  long long int sum = 0;
  sort(0, N - 1);
  if (cod > cev) {
    f = 1;
    cod = cod - cev - 1;
  } else
    cod = cev - cod - 1;
  c = 0;
  for (i = 0; i < N; i++) {
    if (arr[i] % 2 == f) {
      if (c < cod) sum += arr[i];
      c++;
    }
  }
  printf("%lld\n", sum);
  return 0;
}
