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>
const int MaxN = 2e3 + 10;
int n, od, ev;
int a[MaxN], odd[MaxN], even[MaxN];
int cmp(int a, int b) { return a > b; }
inline int solve1() {
  int ret = 0;
  int cnt = 1;
  ret = odd[1];
  while (cnt <= od && cnt <= ev) {
    ret += odd[cnt + 1] + even[cnt];
    ++cnt;
  }
  return ret;
}
inline int solve2() {
  int ret = 0;
  int cnt = 1;
  ret = even[1];
  while (cnt <= od && cnt <= ev) {
    ret += odd[cnt] + even[cnt + 1];
    ++cnt;
  }
  return ret;
}
int main() {
  int sum = 0;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]), sum += a[i];
  std::sort(a + 1, a + n + 1, cmp);
  for (int i = 1; i <= n; i++) {
    if (a[i] % 2)
      odd[++od] = a[i];
    else
      even[++ev] = a[i];
  }
  int ans1 = solve1();
  int ans2 = solve2();
  printf("%d\n", sum - ((ans1 < ans2) ? ans2 : ans1));
  return 0;
}
