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;
using namespace std::chrono;
const long long N = 5100, MAXN = 1e9 + 10, M = 1e18, LOG = 22, mod = 1e9 + 7,
                mod2 = 1e9 + 9, mod3 = 998244353;
void sieve(int n) {
  vector<int> prime(n + 1, 1);
  prime[0] = prime[1] = 0;
  for (int i = 2; i <= n; ++i) {
    if (!prime[i] || i * 1ll * i > n) continue;
    for (int j = i * i; j <= n; j += i) prime[j] = 0;
  }
}
int n;
int cap[N][N], cflow[N][N];
int parent[N];
vector<int> adj[N];
int st, en;
int spfa() {
  int dist[n + 3], state[n + 3];
  memset(state, 0, sizeof(state));
  for (int i = 0; i < n + 3; i++) dist[i] = MAXN;
  dist[st] = 0;
  queue<int> pq;
  pq.push(st);
  bool succ = 0;
  while (!pq.empty()) {
    int u = pq.front();
    pq.pop();
    if (u == en) succ = 1;
    state[u] = 0;
    for (auto v : adj[u]) {
      int weight;
      if (u == st || u == en || v == st || v == en)
        weight = 0;
      else if (cflow[v][u] > 0)
        weight = -1;
      else
        weight = 1;
      if (dist[u] + weight < dist[v] && cap[u][v] - cflow[u][v] > 0) {
        dist[v] = dist[u] + weight;
        parent[v] = u;
        if (!state[v]) {
          pq.push(v);
          state[v] = 1;
        }
      }
    }
  }
  return (succ ? dist[en] : 0);
}
int maxflow(int s, int t) {
  long long flow = 0, cost = 0;
  int new_flow;
  while (new_flow = spfa()) {
    int cur = t;
    flow++;
    while (cur != s) {
      int prev = parent[cur];
      if (cap[prev][cur] != MAXN) cflow[prev][cur]++;
      if (cap[cur][prev] != MAXN) cflow[cur][prev]--;
      cur = prev;
    }
    cost += new_flow;
  }
  return cost;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  cin >> n;
  st = n, en = n + 1;
  for (int i = 0; i < n; i++) {
    int d;
    cin >> d;
    if (d == 1) {
      adj[st].push_back(i);
      adj[i].push_back(st);
      cap[st][i] = 1;
    } else {
      adj[en].push_back(i);
      adj[i].push_back(en);
      cap[i][en] = 1;
    }
    if (i != 0) {
      adj[i].push_back(i - 1);
      adj[i - 1].push_back(i);
      cap[i][i - 1] = MAXN;
      cap[i - 1][i] = MAXN;
    }
  }
  cout << maxflow(st, en) << endl;
  return 0;
}
