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>
#define f first
#define s second
#define fore(i,a,b) for(int i = (a), ThxMK = (b); i < ThxMK; ++i)
#define pb push_back
#define all(s) begin(s), end(s)
#define _ ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define sz(s) int(s.size())
#define ENDL '\n'
using namespace std;
typedef long double ld;
typedef long long lli;
typedef pair<lli,lli> ii;
typedef vector<lli> vi;
#define deb(x) cout << #x": " << (x) << endl;




void solve() {
  lli n; cin >> n;
  vi v(n);
  vi p;
  fore(i, 0, n) {
    cin >> v[i];
    if(v[i]) p.pb(i);
  }
  vi mn(n, 1e16);
  mn[0] = 0;
  int cnt = 1, m = sz(p);

  fore(i, 0, n) {
    if(!v[i]) {
      for(int j = min(m, cnt) - 1; j >= 0; --j) {
        mn[j + 1] = min(mn[j + 1], mn[j] + abs(p[j] - i));
      }
      ++cnt;
    }
  }
  cout << mn[m] << ENDL;
}

int main(){ _
  solve();
  return 0;
}
