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
import heapq


def solve(numbers, parity):
    # split into even and odd numbers
    even_numbers = [-n for n in numbers if n % 2 == 0]
    heapq.heapify(even_numbers)
    odd_numbers = [-n for n in numbers if n % 2 == 1]
    heapq.heapify(odd_numbers)

    def remove_element(some_numbers):
        if len(some_numbers) == 0:
            return True
        heapq.heappop(some_numbers)
        return False

    stop = False
    while not stop:
        if parity == 0:
            stop = remove_element(even_numbers)
        elif parity == 1:
            stop = remove_element(odd_numbers)

        parity = 1 - parity

    # compute sum
    return - sum(even_numbers) - sum(odd_numbers)


def main():
    n = int(input().strip())
    numbers = list(map(int, input().strip().split()))

    even_r = solve(numbers, 0)
    odd_r = solve(numbers, 1)

    print(min(even_r, odd_r))

if __name__ == "__main__":
    main()
