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.
// Problem ID - 1525D

#define debug(...)
#include <bits/stdc++.h>
#include <climits>
using namespace std;
void solve();

#define ll long long

const int MAX_N = 100001;
const int MOD = 1000*1000*1000 + 7;
const int INF = 1000*1000*1000;
const ll LINF = 1ll*1000000*1000000*1000000;

int main() {

    ios_base::sync_with_stdio(false);
    cin.tie(NULL);

    int t=1;
    //cin >> t;
    for (int i = 0; i < t; ++i) {
        solve();
        cout << "\n";
    }

    cerr << "time taken : " << (float)clock()/CLOCKS_PER_SEC << " secs" << "\n";
    return 0;
}

void solve() {

	int n; cin >> n;

	vector<int> z, o;

	for (int i = 0; i < n; ++i){
		int a; cin >> a;
		a == 0? z.push_back(i): o.push_back(i);
	}

	sort(z.begin(), z.end());
	sort(o.begin(), o.end());

	debug(o, z);

	int x = o.size(), y = z.size();

	vector<vector<int>> dp(x+1, vector<int>(y+1, INF));
	dp[0].assign(y+1, 0);

	debug(dp);

	for (int i = 1; i < x+1; ++i){
		for (int j = 1; j < y+1; ++j){
			dp[i][j] = min(dp[i-1][j-1] + abs(o[i-1] - z[j-1]), dp[i][j-1]);
			debug(dp);
		}
	} 

	cout << dp[x][y];
}