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>
int odd[2005];
int even[2005];
bool cmp(int a, int b) { return a < b; }
using namespace std;
int main() {
  int n, t, leno = 1, lene = 1;
  int i;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    scanf("%d", &t);
    if (t & 1) {
      even[lene] = t;
      lene++;
    } else {
      odd[leno] = t;
      leno++;
    }
  }
  lene--;
  leno--;
  if (abs(lene - leno) == 1 || abs(lene - leno) == 0) {
    printf("0");
    return 0;
  }
  sort(odd + 1, odd + 1 + leno, cmp);
  sort(even + 1, even + 1 + lene, cmp);
  int sum = 0;
  for (i = 1; i <= abs(leno - lene) - 1; i++) {
    if (leno > lene)
      sum += odd[i];
    else
      sum += even[i];
  }
  printf("%d", sum);
  return 0;
}
