B. Parity Alternated Deletions

Polycarp has an array aa consisting of nn 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−1n−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 nn (1≤n≤20001≤n≤2000) — the number of elements of aa.

The second line of the input contains nn integers a1,a2,…,ana1,a2,…,an (0≤ai≤1060≤ai≤106), where aiai is the ii-th element of aa.

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

代码：

 

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define inf 0x3f3f3f3f
#define mem(a,b) memset(a,b,sizeof(a))

int a[2005],b[2005]; 

int main()
{
	int n,i=0,j=0,k;
	cin>>n;
	while(n--)
	{
		int t;
		cin>>t;
		if(t%2==1)
			a[i++]=t;
		else
			b[j++]=t;
	}
	sort(a,a+i);
	sort(b,b+j);
//	for(k=0;k<i;k++)
//		cout<<a[k]<<" ";
//	cout<<endl;
//	for(k=0;k<j;k++)
//		cout<<b[k]<<" ";
	int s=0;
	if(i>j)
	{
		i--;
		for(k=0;k<i-j;k++)
			s+=a[k];
	}
	else if(j>i)
	{
		j--;
		for(k=0;k<j-i;k++)
			s+=b[k];
	}
	else
		s=0;
	cout<<s<<endl;
	return 0;
}