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;
#pragma warning(disable : 4996)
long long n, a[1 << 18], sum;
vector<long long> v1, v2;
int main() {
  scanf("%lld", &n);
  for (int i = 0; i < n; i++) {
    scanf("%lld", &a[i]);
    sum += a[i];
    if (a[i] % 2 == 0)
      v1.push_back(a[i]);
    else
      v2.push_back(a[i]);
  }
  sort(v1.begin(), v1.end());
  sort(v2.begin(), v2.end());
  int c1 = v1.size() - 1, c2 = v2.size() - 1, ty = 0;
  if (v1.size() < v2.size()) ty = 1;
  while (true) {
    if (ty == 0) {
      if (c1 < 0) break;
      sum -= v1[c1];
      c1--;
    }
    if (ty == 1) {
      if (c2 < 0) break;
      sum -= v2[c2];
      c2--;
    }
    ty ^= 1;
  }
  printf("%lld\n", sum);
  return 0;
}
