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
def MinimizeleftOverSum(a, n) : 
      
    v1, v2 = [], [];  
    for i in range(n) :  
          
        if (a[i] % 2) : 
            v1.append(a[i]);  
        else : 
            v2.append(a[i]);  
      
    # If more odd elements  
    if (len(v1) > len(v2)) :  
  
        # Sort the elements  
        v1.sort();  
        v2.sort();  
  
        # Left-over elements  
        x = len(v1) - len(v2) - 1;  
  
        sum = 0;  
        i = 0;  
  
        # Find the sum of leftover elements  
        while (i < x) :  
            sum += v1[i]; 
            i += 1
  
        # Return the sum  
        return sum;  
      
    # If more even elements  
    elif (len(v2) > len(v1)) : 
  
        # Sort the elements  
        v1.sort();  
        v2.sort();  
  
        # Left-over elements  
        x = len(v2) - len(v1) - 1;  
  
        sum = 0;  
        i = 0;  
  
        # Find the sum of leftover elements  
        while (i < x) : 
            sum += v2[i]; 
            i += 1
          
        # Return the sum  
        return sum;  
      
    # If same elements  
    else : 
        return 0;  
  
# Driver code  
if __name__ == "__main__" :  
  
    n = int(input())
    a = list(map(int,input().split()))
    n = len(a);  
      
    print(MinimizeleftOverSum(a, n));  