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 long long int p_mod = 9999999999999983;
const long double pi = 3.14159265358979323;
const long long int N = 1e6 + 9;
const long long int mod = 1e9 + 7;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
int rand(int l, int r) {
  uniform_int_distribution<int> uid(l, r);
  return uid(rng);
}
long long int powerk(long long int x, long long int y);
long long int a[N];
void solve() {
  long long int n, m = 0, flag = 0, t = 0;
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
    t += a[i];
    if (a[i] & 1) m++;
  }
  flag = n - m;
  sort(a + 1, a + n + 1);
  if (flag > m)
    flag = m + 1;
  else if (m > flag)
    m = flag + 1;
  long long int o = 0, e = 0, sum = 0;
  for (int i = n; i >= 1; --i) {
    if (a[i] & 1) {
      if (m) sum += a[i], m--;
    } else {
      if (flag) sum += a[i], flag--;
    }
  }
  cout << t - sum;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long int t = 1;
  while (t--) {
    solve();
  }
}
long long int powerk(long long int x, long long int y) {
  if (y == 0) return 1;
  if (y == 1) return x % mod;
  if (y & 1)
    return ((powerk((x * x) % mod, y / 2) % mod) * x) % mod;
  else
    return powerk((x * x) % mod, y / 2) % mod;
}
