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;
void __print(int x) { cerr << x; }
void __print(long x) { cerr << x; }
void __print(long long x) { cerr << x; }
void __print(unsigned x) { cerr << x; }
void __print(unsigned long x) { cerr << x; }
void __print(unsigned long long x) { cerr << x; }
void __print(float x) { cerr << x; }
void __print(double x) { cerr << x; }
void __print(long double x) { cerr << x; }
void __print(char x) { cerr << '\'' << x << '\''; }
void __print(const char *x) { cerr << '\"' << x << '\"'; }
void __print(const string &x) { cerr << '\"' << x << '\"'; }
void __print(bool x) { cerr << (x ? "true" : "false"); }
template <typename T, typename V>
void __print(const pair<T, V> &x) {
  cerr << '{';
  __print(x.first);
  cerr << ',';
  __print(x.second);
  cerr << '}';
}
template <typename T>
void __print(const T &x) {
  int f = 0;
  cerr << '{';
  for (auto &i : x) cerr << (f++ ? "," : ""), __print(i);
  cerr << "}";
}
void _print() { cerr << "]\n"; }
template <typename T, typename... V>
void _print(T t, V... v) {
  __print(t);
  if (sizeof...(v)) cerr << ", ";
  _print(v...);
}
const int maxn = 5005;
int n, a[maxn], zero[maxn], one[maxn];
int L[maxn], R[maxn];
int Cal(int l, int r) {
  if (one[r] - one[l - 1] > zero[r] - zero[l - 1]) return -1;
  set<int> s;
  for (int i = l; i <= r; ++i) {
    if (a[i] == 0) s.insert(i);
  }
  int ans = 0;
  for (int i = l; i <= r; ++i) {
    if (a[i] == 0) continue;
    auto it1 = s.lower_bound(i);
    auto it2 = s.lower_bound(i);
    int cur1 = 2e9, cur2 = 2e9;
    if (it1 != s.end()) cur1 = *it1;
    if (it2 != s.begin()) {
      --it2;
      cur2 = *it2;
    }
    if (abs(cur1 - i) < abs(cur2 - i)) {
      s.erase(it1);
      ans += abs(cur1 - i);
    } else {
      s.erase(it2);
      ans += abs(cur2 - i);
    }
  }
  return ans;
}
void Solve(int T) {
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
    zero[i] = zero[i - 1];
    one[i] = one[i - 1];
    if (a[i] == 0)
      zero[i]++;
    else
      one[i]++;
  }
  int ans = 2e9;
  for (int i = 1; i <= n; ++i) {
    L[i] = Cal(1, i);
    R[i] = Cal(i, n);
  }
  for (int i = 1; i <= n; ++i) {
    if (L[i] == -1) continue;
    for (int j = i + 1; j <= n + 1; ++j) {
      if (one[i] < one[j - 1]) break;
      if (R[j] == -1) continue;
      ans = min(ans, L[i] + R[j]);
    }
  }
  cout << ans;
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
  if (fopen("CF"
            ".inp",
            "r")) {
    freopen(
        "CF"
        ".inp",
        "r", stdin);
    freopen(
        "CF"
        ".out",
        "w", stdout);
  }
  int test = 1;
  for (int i = 1; i <= test; ++i) {
    Solve(i);
  }
}
