There are n armchairs, numbered from 1 to n from left to right. Some armchairs are occupied by people (at most one person per armchair), others are not. The number of occupied armchairs is not greater than n/2.

For some reason, you would like to tell people to move from their armchairs to some other ones. If the i-th armchair is occupied by someone and the j-th armchair is not, you can tell the person sitting in the i-th armchair to move to the j-th armchair. The time it takes a person to move from the i-th armchair to the j-th one is |i - j| minutes. You may perform this operation any number of times, but these operations must be done sequentially, i. e. you cannot tell a person to move until the person you asked to move in the last operation has finished moving to their destination armchair.

You want to achieve the following situation: every seat that was initially occupied must be free. What is the minimum time you need to do it?

Input

The first line contains one integer n (2 ≤ n ≤ 5000) — the number of armchairs.

The second line contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 1). a_i = 1 means that the i-th armchair is initially occupied, a_i = 0 means that it is initially free. The number of occupied armchairs is at most n/2.

Output

Print one integer — the minimum number of minutes you have to spend to achieve the following situation: every seat that was initially occupied must be free.

Examples

Input


7
1 0 0 1 0 0 1


Output


3


Input


6
1 1 1 0 0 0


Output


9


Input


5
0 0 0 0 0


Output


0

Note

In the first test, you can perform the following sequence:

  1. ask a person to move from armchair 1 to armchair 2, it takes 1 minute; 
  2. ask a person to move from armchair 7 to armchair 6, it takes 1 minute; 
  3. ask a person to move from armchair 4 to armchair 5, it takes 1 minute. 



In the second test, you can perform the following sequence:

  1. ask a person to move from armchair 1 to armchair 4, it takes 3 minutes; 
  2. ask a person to move from armchair 2 to armchair 6, it takes 4 minutes; 
  3. ask a person to move from armchair 4 to armchair 5, it takes 1 minute; 
  4. ask a person to move from armchair 3 to armchair 4, it takes 1 minute. 



In the third test, no seat is occupied so your goal is achieved instantly.
#include <bits/stdc++.h>     

using namespace std;

const int INF = int(1e9);

int main()
{
	int n;
	cin >> n;
	vector<int> a(n);
	for(int i = 0; i < n; i++)
		cin >> a[i];
	vector<int> pos;
	for(int i = 0; i < n; i++)
		if(a[i] == 1)
			pos.push_back(i);
	int k = pos.size();
	vector<vector<int>> dp(n + 1, vector<int>(k + 1, INF));
	dp[0][0] = 0;
	for(int i = 0; i < n; i++)
		for(int j = 0; j <= k; j++)
		{
		 	if(dp[i][j] == INF) continue;
		 	dp[i + 1][j] = min(dp[i + 1][j], dp[i][j]);
		 	if(j < k && a[i] == 0)
		 		dp[i + 1][j + 1] = min(dp[i + 1][j + 1], dp[i][j] + abs(pos[j] - i));
		}
	cout << dp[n][k] << endl;
}