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 n, od[2010], ev[2010], cnt1 = 0, cnt2 = 0, sum = 0;
bool cmp(int _, int __) { return _ > __; }
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    int a;
    scanf("%d", &a);
    if (a % 2 != 0)
      od[++cnt1] = a;
    else
      ev[++cnt2] = a;
  }
  sort(od + 1, od + cnt1 + 1, cmp);
  sort(ev + 1, ev + cnt2 + 1, cmp);
  int x = min(cnt1, cnt2);
  int xx = x;
  for (++x; x <= max(cnt1, cnt2); x++) sum += od[x] + ev[x];
  printf("%d", sum - max(od[xx + 1], ev[xx + 1]));
  return 0;
}
