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 int MOD = 1000 * 1000 * 1000 + 7;
const int INF = 1 * 1000 * 1000 * 1000 + 200;
const int MAXN = 5 * 100 * 1000 + 10;
const int MLOG = 30;
const int NMAX = 20;
int main() {
  int n;
  cin >> n;
  int nm[2] = {0, 0}, a;
  vector<int> vc[2];
  for (int i = 0; i < n; i++) cin >> a, nm[a % 2]++, vc[a % 2].push_back(a);
  int h = 1;
  if (nm[0] > nm[1]) swap(nm[1], nm[0]), h = 0;
  int all = max(0, nm[1] - nm[0] - 1), ans = 0;
  sort(vc[h].begin(), vc[h].end());
  for (int i = 0; i < all; i++) ans += vc[h][i];
  cout << ans;
}
