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;
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
#define mp            make_pair
#define pb            push_back
#define ppb           pop_back
#define all(x)        (x).begin(),(x).end()
#define int           long long
#define double        long double
#define rep(i,a,b)    for(int i=a;i<=b;i++)
#define rrep(i,b,a)    for(int i=b;i>=a;i--)

#ifndef ONLINE_JUDGE
#define debug(x) cerr << #x <<" "; _print(x); cerr << endl;
#else
#define debug(x)
#endif

void _print(int t) {cerr << t;}
void _print(string t) {cerr << t;}
void _print(char t) {cerr << t;}
void _print(double t) {cerr << t;}
void _print(bool t) {cerr << t;}

template <class T, class V> void _print(pair <T, V> p);
template <class T> void _print(vector <T> v);
template <class T> void _print(set <T> v);
template <class T, class V> void _print(map <T, V> v);
template <class T> void _print(multiset <T> v);
template <class T, class V> void _print(pair <T, V> p) {cerr << "{"; _print(p.first); cerr << ","; _print(p.second); cerr << "}";}
template <class T> void _print(vector <T> v) {cerr << "[ "; for (T i : v) {_print(i); cerr << " ";} cerr << "]";}
template <class T> void _print(set <T> v) {cerr << "[ "; for (T i : v) {_print(i); cerr << " ";} cerr << "]";}
template <class T> void _print(multiset <T> v) {cerr << "[ "; for (T i : v) {_print(i); cerr << " ";} cerr << "]";}
template <class T, class V> void _print(map <T, V> v) {cerr << "[ "; for (auto i : v) {_print(i); cerr << " ";} cerr << "]";}

void solve() {

	int n;
	cin >> n;

	std::vector<int> v(n), mt, occ;

	rep(i, 0, n - 1) {
		cin >> v[i];
		if (v[i] == 0) {
			mt.pb(i);
		}
		else {
			occ.pb(i);
		}
	}



	//memset(dp, -1, sizeof dp);
	debug(mt);
	debug(occ);


	int o = occ.size(), m = mt.size();
	vector<vector<int>> dp(o + 1, vector<int> (m + 1, 2e18 + 1));
	dp[0][0] = 0;

	rep(i, 0, m) {
		dp[0][i] = 0;
	}


	for (int i = 1; i <= o; i++) {
		for (int j = 1; j <= m; j++) {
			if (j >= i) {
				//debug(mp(i, j));
				dp[i][j] = min(dp[i - 1][j - 1] + abs(occ[i - 1] - mt[j - 1]), dp[i][j - 1]);
			}
		}
	}
	debug(dp);


	cout << dp[o][m];



}

signed main() {

	ios_base::sync_with_stdio(false);
	cin.tie(0); cout.tie(0);
#ifndef ONLINE_JUDGE
	freopen("input.txt", "r", stdin);
	freopen("output.txt", "w", stdout);
	freopen("Error.txt", "w", stderr);
#endif

	int t = 1;
	//cin >> t;
	cout << setprecision(9) << fixed;

	for (int i = 1; i <= t; i++) {

		solve();
	}

	return 0;
}
//ceil a/b = a+b-1/b
