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 ll = long long;
using ld = long double;
using pll = pair<ll, ll>;
constexpr ll MOD = 1e9 + 7;
//constexpr ll MOD = 998244353;
//constexpr ll MOD = ;
ll mod(ll A, ll M) {return (A % M + M) % M;}
constexpr ll INF = 1LL << 60;
template<class T> bool chmin(T& a, T b) {if (a > b) {a = b; return true;} return false;}
template<class T> bool chmax(T& a, T b) {if (a < b) {a = b; return true;} return false;}
ll divceil(ll A, ll B) {return (A + (B - 1)) / B;}
#define FINALANS(A) do {cout << (A) << '\n'; exit(0);} while (false)

int main()
{
  ll N;
  cin >> N;
  vector<ll> ones, zeros;
  for (ll i = 0; i < N; i++)
  {
    ll a;
    cin >> a;
    if (a == 1)
      ones.push_back(i);
    else
      zeros.push_back(i);
  }

  ll M1 = ones.size(), M0 = zeros.size();

  vector<vector<ll>> dp(M0 + 1, vector<ll>(M1 + 1, INF));
  dp.at(0).at(0) = 0;
  for (ll i = 0; i < M0; i++)
  {
    for (ll j = 0; j <= M1; j++)
    {
      if (j < M1)
        chmin(dp.at(i + 1).at(j + 1), dp.at(i).at(j) + abs(zeros.at(i) - ones.at(j)));
      chmin(dp.at(i + 1).at(j), dp.at(i).at(j));
    }
  }

  ll ans = dp.at(M0).at(M1);
  cout << ans << '\n';
}