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>

#ifndef ATCODER_MINCOSTFLOW_HPP
#define ATCODER_MINCOSTFLOW_HPP 1

#include <algorithm>
#include <cassert>
#include <limits>
#include <queue>
#include <vector>

namespace atcoder {

template <class Cap, class Cost> struct mcf_graph {
  public:
    mcf_graph() {}
    mcf_graph(int n) : _n(n), g(n) {}

    int add_edge(int from, int to, Cap cap, Cost cost) {
        assert(0 <= from && from < _n);
        assert(0 <= to && to < _n);
        int m = int(pos.size());
        pos.push_back({from, int(g[from].size())});
        int from_id = int(g[from].size());
        int to_id = int(g[to].size());
        if (from == to) to_id++;
        g[from].push_back(_edge{to, to_id, cap, cost});
        g[to].push_back(_edge{from, from_id, 0, -cost});
        return m;
    }

    struct edge {
        int from, to;
        Cap cap, flow;
        Cost cost;
    };

    edge get_edge(int i) {
        int m = int(pos.size());
        assert(0 <= i && i < m);
        auto _e = g[pos[i].first][pos[i].second];
        auto _re = g[_e.to][_e.rev];
        return edge{
            pos[i].first, _e.to, _e.cap + _re.cap, _re.cap, _e.cost,
        };
    }
    std::vector<edge> edges() {
        int m = int(pos.size());
        std::vector<edge> result(m);
        for (int i = 0; i < m; i++) {
            result[i] = get_edge(i);
        }
        return result;
    }

    std::pair<Cap, Cost> flow(int s, int t) {
        return flow(s, t, std::numeric_limits<Cap>::max());
    }
    std::pair<Cap, Cost> flow(int s, int t, Cap flow_limit) {
        return slope(s, t, flow_limit).back();
    }
    std::vector<std::pair<Cap, Cost>> slope(int s, int t) {
        return slope(s, t, std::numeric_limits<Cap>::max());
    }
    std::vector<std::pair<Cap, Cost>> slope(int s, int t, Cap flow_limit) {
        assert(0 <= s && s < _n);
        assert(0 <= t && t < _n);
        assert(s != t);
        // variants (C = maxcost):
        // -(n-1)C <= dual[s] <= dual[i] <= dual[t] = 0
        // reduced cost (= e.cost + dual[e.from] - dual[e.to]) >= 0 for all edge
        std::vector<Cost> dual(_n, 0), dist(_n);
        std::vector<int> pv(_n), pe(_n);
        std::vector<bool> vis(_n);
        auto dual_ref = [&]() {
            std::fill(dist.begin(), dist.end(),
                      std::numeric_limits<Cost>::max());
            std::fill(pv.begin(), pv.end(), -1);
            std::fill(pe.begin(), pe.end(), -1);
            std::fill(vis.begin(), vis.end(), false);
            struct Q {
                Cost key;
                int to;
                bool operator<(Q r) const { return key > r.key; }
            };
            std::priority_queue<Q> que;
            dist[s] = 0;
            que.push(Q{0, s});
            while (!que.empty()) {
                int v = que.top().to;
                que.pop();
                if (vis[v]) continue;
                vis[v] = true;
                if (v == t) break;
                // dist[v] = shortest(s, v) + dual[s] - dual[v]
                // dist[v] >= 0 (all reduced cost are positive)
                // dist[v] <= (n-1)C
                for (int i = 0; i < int(g[v].size()); i++) {
                    auto e = g[v][i];
                    if (vis[e.to] || !e.cap) continue;
                    // |-dual[e.to] + dual[v]| <= (n-1)C
                    // cost <= C - -(n-1)C + 0 = nC
                    Cost cost = e.cost - dual[e.to] + dual[v];
                    if (dist[e.to] - dist[v] > cost) {
                        dist[e.to] = dist[v] + cost;
                        pv[e.to] = v;
                        pe[e.to] = i;
                        que.push(Q{dist[e.to], e.to});
                    }
                }
            }
            if (!vis[t]) {
                return false;
            }

            for (int v = 0; v < _n; v++) {
                if (!vis[v]) continue;
                // dual[v] = dual[v] - dist[t] + dist[v]
                //         = dual[v] - (shortest(s, t) + dual[s] - dual[t]) + (shortest(s, v) + dual[s] - dual[v])
                //         = - shortest(s, t) + dual[t] + shortest(s, v)
                //         = shortest(s, v) - shortest(s, t) >= 0 - (n-1)C
                dual[v] -= dist[t] - dist[v];
            }
            return true;
        };
        Cap flow = 0;
        Cost cost = 0, prev_cost_per_flow = -1;
        std::vector<std::pair<Cap, Cost>> result;
        result.push_back({flow, cost});
        while (flow < flow_limit) {
            if (!dual_ref()) break;
            Cap c = flow_limit - flow;
            for (int v = t; v != s; v = pv[v]) {
                c = std::min(c, g[pv[v]][pe[v]].cap);
            }
            for (int v = t; v != s; v = pv[v]) {
                auto& e = g[pv[v]][pe[v]];
                e.cap -= c;
                g[v][e.rev].cap += c;
            }
            Cost d = -dual[s];
            flow += c;
            cost += c * d;
            if (prev_cost_per_flow == d) {
                result.pop_back();
            }
            result.push_back({flow, cost});
            prev_cost_per_flow = d;
        }
        return result;
    }

  private:
    int _n;

    struct _edge {
        int to, rev;
        Cap cap;
        Cost cost;
    };

    std::vector<std::pair<int, int>> pos;
    std::vector<std::vector<_edge>> g;
};

}  // namespace atcoder

#endif  // ATCODER_MINCOSTFLOW_HPP


using namespace std;
using namespace atcoder;
using ll = long long;
using ld = long double;
using uint = unsigned int;
using ull = unsigned long long;
const int MOD = 1e9 + 7;

int main() {
  ll mx = 1e9;
  ll n,cnt;
  cin >> n;
  ll m = n + 2;
  ll s = n;
  ll g = s + 1;
  cnt = 0;
  mcf_graph<ll, ll> graph(m+4);
  for (ll i = 0; i < n; i++) {
    ll c;
    cin >> c;
    cnt += c;
    if(c == 0){
        graph.add_edge(s, i, 1LL, 1LL);
    }else{
        graph.add_edge(i, g, 1LL, 1LL);
    }
  }
  for (ll i = 0; i < n; i++) {
    if(i != 0){
        graph.add_edge(i, i-1, n, 1LL);
        graph.add_edge(i-1, i, n, 1LL);
    }else if(i != n-1){
        graph.add_edge(i+1, i, n, 1LL);
        graph.add_edge(i, i+1, n, 1LL);
    }
  }
  auto ans = graph.flow(s,g,cnt);
  ll prt = ans.second-cnt*2;
  cout << prt << endl;
}