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 int long long
#define sz(x) (int)x.size()
#define all(x) (x).begin(), (x).end()
#define pb push_back
#define eb emplace_back
#define ff first
#define ss second
#define pi pair<int, int>

string to_string(const string& s) {return '"' + s + '"';}
string to_string(const char* s) {return to_string((string) s);}
string to_string(bool b) {return (b ? "true" : "false");}
string to_string(vector<bool> v) {bool first = true;string res = "{";for (int i = 0; i < static_cast<int>(v.size()); i++) {if (!first) {res += ", ";}first = false;res += to_string(v[i]);}res += "}";return res;}
template <size_t N> string to_string(bitset<N> v) {string res = "";for (size_t i = 0; i < N; i++) {res += static_cast<char>('0' + v[i]);}return res;}
template <class A, class B> string to_string(pair<A, B> p) {return "(" + to_string(p.first) + ", " + to_string(p.second) + ")";}
template <class A, class B, class C> string to_string(tuple<A, B, C> p) {return "(" + to_string(get<0>(p)) + ", " + to_string(get<1>(p)) + ", " + to_string(get<2>(p)) + ")";}
template <class A, class B, class C, class D> string to_string(tuple<A, B, C, D> p) {return "(" + to_string(get<0>(p)) + ", " + to_string(get<1>(p)) + ", " + to_string(get<2>(p)) + ", " + to_string(get<3>(p)) + ")";}
template <class A> string to_string(A v) {bool first = true;string res = "{";for (const auto &x : v) {if (!first) {res += ", ";}first = false;res += to_string(x);}res += "}";return res;}

template <class Arg1>
void __f(const char* name, Arg1&& arg1){
  cerr << name << " : " << to_string(arg1) << "\n";
}

template <class Arg1, class... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args){
  const char* comma = strchr(names + 1, ',');
  cerr.write(names, comma - names) << " : " << to_string(arg1) <<" | ";
  __f(comma+1, args...);
}

#ifdef LOCAL
#define debug(...) __f(#__VA_ARGS__, __VA_ARGS__)
#else
#define debug(...) 69
#endif

signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int n;
  cin >> n;
  vector<int> a, b;
  for (int i = 0; i < n; i++) {
    int x;
    cin >> x;
    if (x) a.pb(i);
    else b.pb(i);
  }
  vector<vector<int>> dp(sz(a), vector<int>(sz(b) + 1, -1));
  function<int(int, int)> solve = [&](int i, int j) -> int {
    debug(i, j);
    if (i == sz(a)) return 0LL;
    int& cur = dp[i][j];
    if (~cur) return cur;
    cur = (int) 1e15;
    if (j < sz(b)) {
      // take it, or wait
      cur = min(abs(a[i] - b[j]) + solve(i + 1, j + 1), solve(i, j + 1));
    }
    return cur;
  };
  cout << solve(0, 0) << "\n";
  return 0;
}