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;
int n, x, ans;
vector<bool> v;
vector<int> w;
void del(int i) {
  vector<bool> vv;
  for (int j = v.size() - 1; j > i + 1; --j) {
    vv.push_back(v[j]);
    v.pop_back();
  }
  v.pop_back();
  v.pop_back();
  while (!vv.empty()) {
    v.push_back(vv[vv.size() - 1]);
    vv.pop_back();
  }
  if (i > 0) w[i - 1] += w[i] + w[i + 1];
  vector<int> ww;
  for (int j = w.size() - 1; j > i + 1; --j) {
    ww.push_back(w[j]);
    w.pop_back();
  }
  w.pop_back();
  w.pop_back();
  while (!ww.empty()) {
    w.push_back(ww[ww.size() - 1]);
    ww.pop_back();
  }
}
void sol() {
  int a, b;
  for (int i = 0; i < v.size() - 1; ++i) {
    if (v[i] ^ v[i + 1]) {
      a = i;
      break;
    }
  }
  b = a;
  for (int i = a + 1; i < v.size() - 1; ++i) {
    if (v[i] ^ v[i + 1]) {
      b = i;
      break;
    }
  }
  if (w[a] > w[b] || v[0] == 1) {
    ans += w[a];
    del(a);
  } else {
    ans += w[b];
    del(b);
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  bool bl;
  while (n--) {
    cin >> bl;
    v.push_back(bl);
    w.push_back(1);
    if (bl) ++x;
  }
  while (x--) sol();
  cout << ans;
}
