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.
#pragma GCC target ("avx2")
#pragma GCC optimization ("O3")
#pragma GCC optimization ("unroll-loops")
#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define int long long
#define all(x) x.begin() , x.end()
#define ld long double
const long long inf = 1e18 + 5LL;
const int mod = 1e9 + 7LL;
const int N = (2e5 + 10);
void solve(int t);
void ITO();
int32_t main() {
	ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);
	ITO();
	int t = 1;
	// cin >> t;
	for (int i = 0; i < t; i++) solve(i + 1);
	return 0;
}
void solve(int TcNumber)
{
	int n; cin >> n;
	int arr[n]; for (auto &x : arr) cin >> x;
	vector<int> o , z;
	for (int i = 0; i < n; i++) {
		if (arr[i]) o.push_back(i);
		else z.push_back(i);
	}

	n = o.size();
	int m = z.size();
	int dp[n + 1][m + 1];
	for (int i = 0; i <= n; i++) {
		for (int j = 0; j <= m; j++) {
			if (i == 0) dp[i][j] = 0;
			else if (j == 0) dp[i][j] = inf;
			else {
				dp[i][j] = min({
					dp[i][j - 1] ,
					abs(o[i - 1] - z[j - 1]) + dp[i - 1][j - 1]
				});
			}
		}
	}
	cout << dp[n][m];
	cout << "\n";
	return ;
}
void ITO()
{
#ifndef ONLINE_JUDGE
	freopen("inputf.in", "r", stdin);
	freopen("outputf1.in", "w", stdout);
#endif
}
