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 int mod = 1e9 + 7;
long long binpow(long long a, long long b) {
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = res * a;
    a = a * a;
    b >>= 1;
  }
  return res;
}
int countSetBits(int n) {
  unsigned int count = 0;
  while (n) {
    n &= (n - 1);
    count++;
  }
  return count;
}
int powerOfTwo(int n) { return n && (!(n & (n - 1))); }
void solve() {
  long long int n;
  cin >> n;
  vector<long long int> a(n);
  for (long long int i = 0; i < n; i++) cin >> a[i];
  sort(a.rbegin(), a.rend());
  vector<long long int> o;
  vector<long long int> e;
  long long int even = 0;
  long long int odd = 0;
  for (long long int i = 0; i < n; i++) {
    if (a[i] % 2 == 0) {
      even++;
      e.push_back(a[i]);
    } else {
      odd++;
      o.push_back(a[i]);
    }
  }
  long long int sum = 0;
  if (even > odd) {
    for (long long int i = odd + 1; i < even; i++) sum = sum + e[i];
  } else if (odd > even) {
    for (long long int i = even + 1; i < odd; i++) sum = sum + o[i];
  }
  cout << sum;
}
int main() {
  solve();
  return 0;
}
