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 <iostream>
#include <vector>
#include <algorithm>
#include <set>
#include <map>
#include <string>
#include <stdio.h>
#include <unordered_set>
#include <deque>
#include <unordered_map>
#include <ctime>
#include <fstream>
#include <iomanip>
#include <cmath>
#define int long long
using namespace std;
signed main() {
	ios_base::sync_with_stdio(false);
	srand(time(NULL));
	cout.tie(0);
	cin.tie(0);
	int n;
	cin >> n;
	vector <vector <int> > dp(n+1, vector <int>(n+1, 1e17));
	dp[0][0] = 0;
	for (int i = 0; i <= n; ++i) {
		dp[0][i] = 0;
	}
	vector <int> a(n);
	for (int i = 0; i < n; ++i) {
		cin >> a[i];
	}
	for (int i = 1; i <= n; ++i) {
		if (a[i - 1] == 0) {
			dp[i] = dp[i - 1];
		}
		else {
			vector <int> mc(n + 1, 1e17);
			int mx = 1e17;
			for (int z = 0; z <= n; ++z) {
				mc[z] = mx;
				mx = min(mx, dp[i - 1][z]);
			}
			for (int z = 1; z <= n; ++z) {
				if (a[z - 1] == 0) {
					dp[i][z] = mc[z] + abs(z - i);
				}
			}
		}
	}
	int ans = 1e17;
	for (int i = 0; i <= n; ++i) {
		ans = min(dp.back()[i], ans);
	}
	cout << ans;
}
/*
1
7 8
6 1 7 2 3 5 4
R L R L L L L
*/