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>
void quick_sort(int[], int, int);
int partition(int[], int, int);
void quick_sort(int a[], int l, int u) {
  int j;
  if (l < u) {
    j = partition(a, l, u);
    quick_sort(a, l, j - 1);
    quick_sort(a, j + 1, u);
  }
}
int partition(int a[], int l, int u) {
  int v, i, j, temp;
  v = a[l];
  i = l;
  j = u + 1;
  do {
    do i++;
    while (a[i] < v && i <= u);
    do j--;
    while (v < a[j]);
    if (i < j) {
      temp = a[i];
      a[i] = a[j];
      a[j] = temp;
    }
  } while (i < j);
  a[l] = a[j];
  a[j] = v;
  return (j);
}
int main() {
  int c1, c2, n, i, k1 = 0, k2 = 0;
  scanf("%d", &n);
  int a[n];
  for (i = 0; i < n; i++) {
    scanf("%d", &a[i]);
  }
  int save = 0;
  int x[n], y[n];
  for (i = 0; i < n; i++) {
    if (a[i] % 2 == 1)
      x[k1++] = a[i];
    else
      y[k2++] = a[i];
  }
  int min;
  min = k1;
  if (k2 < k1) min = k2;
  quick_sort(x, 0, k1 - 1);
  quick_sort(y, 0, k2 - 1);
  if (min != 0) {
    for (i = 0; i < min; i++) {
      x[k1 - 1 - i] = 0;
      y[k2 - 1 - i] = 0;
    }
    if (k1 > k2) {
      x[k1 - min - 1] = 0;
    } else
      y[k2 - min - 1] = 0;
    int sum1 = 0;
    int sum2 = 0;
    for (i = 0; i < k1; i++) sum1 += x[i];
    for (i = 0; i < k2; i++) sum2 += y[i];
    int ans = sum1 + sum2;
    printf("%d", ans);
  } else {
    if (k1 == 0)
      for (i = 0; i < k2 - 1; i++) save += y[i];
    else
      for (i = 0; i < k1 - 1; i++) save += x[i];
    printf("%d", save);
  }
}
