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;
#define rep(i, n) for (int i = 0; (i) < (int) (n); (i) ++)
#define all(a) (a).begin(), (a).end()
#define rall(a) (a).rbegin(), (a).rend()
#define dump(x) cerr << #x << " = " << x << '\n';
using ll = long long;
template<typename T, typename U> ostream& operator << (ostream& os, const pair<T, U>& p) { os << p.first << " " << p.second; return os; }
template<typename T> ostream& operator << (ostream& os, const vector<T>& vec) { for (int i = 0; i < vec.size(); i ++) { os << vec[i] << (i + 1 == vec.size() ? "" : " "); } return os; }
template<typename T> ostream& operator << (ostream& os, const set<T> &p) { os << "{"; for (auto it = p.begin(); it != p.end(); it ++) { if (it != p.begin()) { os << ", "; } os << *it; } os << "}"; return os; }
template<typename T, typename U> ostream& operator << (ostream& os, const map<T, U> &p) { for (auto it = p.begin(); it != p.end(); it ++) { if (it != p.begin()) { os << ", "; } os << "{" << (*it).first << ": " << (*it).second << "}"; } return os; }
template<typename T> bool chmin(T &a, const T& b) { if (a > b) { a = b; return true; } return false; }
template<typename T> bool chmax(T &a, const T& b) { if (a < b) { a = b; return true; } return false; }

void solve() {
        int n;
        cin >> n;
        vector<int> a(n);
        rep(i, n) {
                cin >> a[i];
        }
        vector<int> zi, oi;
        rep(i, n) {
                if (a[i] == 0) {
                        zi.push_back(i);
                } else {
                        oi.push_back(i);
                }
        }
        const int inf = 1e9;
        vector<vector<int>> dp(oi.size() + 1, vector<int> (zi.size() + 1, inf));
        rep(i, zi.size() + 1) {
                dp[0][i] = 0;
        }
        rep(i, oi.size()) {
                rep(j, zi.size()) {
                        int d = abs(oi[i] - zi[j]);
                        chmin(dp[i + 1][j + 1], dp[i + 1][j]);
                        chmin(dp[i + 1][j + 1], dp[i][j] + d);
                }
        }
        int ans = inf;
        rep(i, zi.size() + 1) {
                chmin(ans, dp[oi.size()][i]);
        }
        cout << ans << '\n';
}

int main() {
        cin.tie(nullptr);
        ios::sync_with_stdio(false);
        // int t; cin >> t;
        int t = 1;
        while (t --) {
                solve();
        }
        return 0;
}
