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;
const double PI = 3.141592653589793238460;
const long long mod = 1e9 + 7;
void FAST() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
}
bool comp(const pair<int, int> &a, const pair<int, int> &b) {
  return (a.first > b.second);
}
void solve() {
  int n;
  cin >> n;
  vector<int> even, odd, sol;
  for (__typeof(n) i = (0) - (0 > n); i != (n) - (0 > n);
       i += 1 - 2 * (0 > n)) {
    int x;
    cin >> x;
    if (x & 1)
      odd.push_back(x);
    else
      even.push_back(x);
  }
  sort(odd.begin(), odd.end());
  sort(even.begin(), even.end());
  int x = min(even.size(), odd.size());
  while (x--) {
    odd.pop_back();
    even.pop_back();
  }
  if (odd.size() > even.size())
    odd.pop_back();
  else if (odd.size() < even.size())
    even.pop_back();
  int sum = 0;
  for (auto x : even) sum += x;
  for (auto x : odd) sum += x;
  cout << sum << '\n';
}
int main() {
  FAST();
  int t = 1;
  while (t--) {
    solve();
  }
  return 0;
}
