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;
vector<int> myodd, myeven;
bool cmp(int a, int b) { return a > b; }
int main() {
  scanf("%d", &n);
  int tmp;
  int sum = 0;
  for (int i = 0; i < n; ++i) {
    scanf("%d", &tmp);
    sum += tmp;
    if (tmp % 2 == 0) {
      myeven.push_back(tmp);
    } else {
      myodd.push_back(tmp);
    }
  }
  sort(myeven.begin(), myeven.end(), cmp);
  sort(myodd.begin(), myodd.end(), cmp);
  int ans1 = 0;
  int len1 = myeven.size(), len2 = myodd.size();
  int i = 0, j = 0;
  for (; i < len1 && j < len2; ++i, ++j) {
    ans1 += myeven[i];
    ans1 += myodd[j];
  }
  if (i < len1) {
    ans1 += myeven[i];
  }
  if (j < len2) {
    ans1 += myodd[j];
  }
  int ans2 = 0;
  i = 0;
  j = 0;
  for (; i < len1 && j < len2; ++i, ++j) {
    ans2 += myodd[j];
    ans2 += myeven[i];
  }
  if (j < len2) {
    ans2 += myodd[j];
  }
  if (i < len1) {
    ans2 += myeven[i];
  }
  printf("%d\n", sum - max(ans1, ans2));
  return 0;
}
