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>
#define sz(x) (int)x.size()
#define chmax(x, v) x = max(x, v)
#define chmin(x, v) x = min(x, v)
#define int long long
#define all(x) x.begin(), x.end()
#define pb push_back
#define pii pair<int, int>
using namespace std;

const int N = 5005, OO = 1e9;
int dp[N][N];

void solve(){
	int nVals; cin >> nVals;
	vector<int> vals(nVals);
	int dernUn[nVals], dernZero[nVals];
	int ind = 0;
	bool tout0 = true;
	for (int& val : vals){
		cin >> val;
		tout0 &= (val == 0);
		if (ind == 0)
			dernZero[ind] = dernUn[ind] = -1;
		else {
			dernUn[ind] = dernUn[ind - 1];
			dernZero[ind] = dernZero[ind - 1];
			if (vals[ind - 1] == 0)
				dernZero[ind] = ind - 1;
			else
				dernUn[ind] = ind - 1;
		}

		ind++;
	}

	if (tout0){
		cout << 0 << endl;
		return;
	}

	int fin = dernUn[nVals - 1];
	if (vals[nVals - 1] == 1)
		fin = nVals - 1;

	for (int i = 0; i < nVals; ++i)
		for (int j = 0; j < nVals; ++j)
			dp[i][j] = OO;
	for (int zero = 0; zero < nVals; ++zero){
		if (vals[zero] != 0) continue;
		for (int un = 0; un < nVals; ++un){
			if (vals[un] != 1) continue;
			if (dernUn[un] == -1){
				dp[zero][un] = abs(zero - un);
			}
			else if (dernZero[zero] != -1) 
				dp[zero][un] = dp[dernZero[zero]][dernUn[un]] + abs(zero - un);

			if (dernZero[zero] != -1)
				chmin(dp[zero][un], dp[dernZero[zero]][un]);
		}
	}
	int mini = OO;
	for (int autre = 0; autre < nVals; ++autre){
		if (vals[autre] == 1) continue;
		chmin(mini, dp[autre][fin]);
	}
	cout << mini << endl;
}

signed main(){
	ios::sync_with_stdio(false);
	cin.tie(nullptr);
	cout.tie(nullptr);

	int t = 1; //cin >> t;
	while (t--)
		solve();
	return 0;
}