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;
void intinput(vector<int> &arr, int size);
int findint(vector<int> &arr, int val);
int findstring(vector<string> &arr, string val);
void sortinc(vector<int> &arr);
void sortdec(vector<int> &arr);
int sum(vector<int> &arr);
void print(vector<int> arr);
int getIndex(vector<int> v, int K);

int dp[5001][5001];
vector<int> op; vector<int> ep;

int recurse(int i, int j, int n, int m)
{
	if (i == n) return 0;

	int av = m - j;
	int re = n - i;
	if (av < re) return 100000000;

	if (dp[i][j] != -1) return dp[i][j];

	int ans1 = recurse(i, j+1, n, m);
	int ans2 = recurse(i+1, j+1, n, m) + abs(op[i] - ep[j]);
	dp[i][j] = min(ans1, ans2);
	return min(ans1, ans2);
	
}


int main()
{

	int n; cin>>n;
	int num;
	op.clear();
	ep.clear();
	for (int i = 0; i < n; i++)
	{
		cin>>num;
		if (num == 0) ep.push_back(i);
		else op.push_back(i);
	}
	int n1 = op.size();
	int n2 = ep.size();

	if (n1 == 0)
	{
		cout<<0<<"\n";
	}
	else
	{
		for (int i = 0; i <= n1; i++)
		{
			for (int j = 0; j <= n2; j++)
			{
				dp[i][j] = -1;
			}
		}
		cout<<recurse(0, 0, n1, n2);
	}
	return 0;
}
















/* Kaam Ka Kachra 
----------------------------------------------------------------------------------------*/
void sortinc(vector<int> &arr) {return sort(arr.begin(), arr.end());}
void sortdec(vector<int> &arr) {return sort(arr.begin(), arr.end(), greater<int>());}
void intinput(vector<int> &arr, int size){int v;while(size--){scanf("%d", &v);arr.push_back(v);}}
int findint(vector<int> &arr, int val){if(find(arr.begin(), arr.end(), val) != arr.end()){return 1;}else{return 0;}}   
int findstring(vector<string> &arr, string val){if(find(arr.begin(), arr.end(), val) != arr.end()){return 1;}else{return 0;}}
int sum(vector<int> &arr){int res = 0;for (int i = 0; i < arr.size(); ++i){res += arr[i];}return res;}
void print(vector<int> arr){for (int i = 0; i < arr.size();i++){cout<<arr[i]<<" ";}cout<<endl;}
int getIndex(vector<int> v, int K){auto it = find(v.begin(), v.end(), K);if (it != v.end()){int index = it - v.begin();return index;}else{return -1;}}