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;
long long n;
void solve() {
  cin >> n;
  long long ocount = 0, ecount = 0;
  long long mineve = LONG_LONG_MAX, minodd = LONG_LONG_MAX;
  vector<long long> oarr, earr;
  for (long long i = 0; i <= n - 1; i++) {
    long long temp;
    cin >> temp;
    if (temp % 2 == 0) {
      earr.push_back(temp);
    }
    if (temp % 2 == 1) {
      oarr.push_back(temp);
    }
  }
  sort(oarr.begin(), oarr.end());
  sort(earr.begin(), earr.end());
  if (oarr.size() == earr.size() + 1 || earr.size() == oarr.size() + 1 ||
      oarr.size() == earr.size()) {
    cout << 0 << '\n';
    return;
  }
  long long ans = 0;
  if (oarr.size() > earr.size()) {
    for (long long i = 1; i <= oarr.size() - earr.size() - 1; i++) {
      ans += oarr[i - 1];
    }
  } else {
    for (long long i = 1; i <= earr.size() - oarr.size() - 1; i++) {
      ans += earr[i - 1];
    }
  }
  cout << max(ans, ans);
}
int main() {
  ios::sync_with_stdio(false);
  solve();
}
