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;
const int64_t MOD = 1e9 + 7;
const int64_t N = 1e6 + 10;
const double PI = 3.141592653589;
const int64_t inf = 1e18;
bool isQuery = false;
bool isValidParenthesis(string s) {
  int64_t c = 0;
  for (int64_t i = 0; i < s.size(); i++) {
    if (s[i] == '(')
      c++;
    else
      c--;
    if (c < 0) return false;
  }
  return c == 0;
}
int64_t longestPrefixSuffix(string s) {
  int64_t n = s.length();
  int64_t lps[n], len = 0, i = 1;
  lps[0] = 0;
  while (i < n) {
    if (s[i] == s[len]) {
      len++;
      lps[i] = len;
      i++;
    } else {
      if (len)
        len = lps[len - 1];
      else
        lps[i++] = 0;
    }
  }
  return lps[n - 1];
}
int64_t power(int64_t x, int64_t n) {
  if (n == 0)
    return 1;
  else if (n % 2 == 0)
    return power(x, n / 2) * power(x, n / 2);
  else
    return x * power(x, n / 2) * power(x, n / 2);
}
int64_t powerMod(int64_t x, int64_t n, int64_t mod) {
  if (n == 0)
    return 1;
  else if (n % 2 == 0)
    return (powerMod(x, n / 2, mod) * powerMod(x, n / 2, mod)) % mod;
  else
    return (x * powerMod(x, n / 2, mod) * powerMod(x, n / 2, mod)) % mod;
}
void solve() {
  int64_t n, c = 0;
  cin >> n;
  vector<int64_t> v(n), mp(n, 0);
  vector<int64_t> one;
  for (int64_t i = 0; i < n; i++) {
    cin >> v[i];
    if (v[i] == 1) one.push_back(i), mp[i] = 1, c++;
  }
  int64_t ans = 0;
  for (auto id : one) {
    vector<int64_t> left(2, 0), right(2, 0);
    for (int64_t i = 0; i < id; i++) {
      if (mp[i] == 1 && v[i] == 0) continue;
      left[v[i]]++;
    }
    int64_t left_o = -1, right_o = -1;
    if (left[0] > left[1]) {
      for (int64_t i = id - 1; i >= 0; i--) {
        if (v[i] == 0 && mp[i] == 0) {
          left_o = i;
          if (n % 2 == 0 && c != n / 2 || n & 1) break;
        }
      }
    }
    for (int64_t i = id + 1; i < n; i++) {
      if (mp[i] == 1 && v[i] == 0) continue;
      right[v[i]]++;
    }
    if (right[0] > right[1]) {
      for (int64_t i = id + 1; i < n; i++) {
        if (v[i] == 0 && mp[i] == 0) {
          right_o = i;
          if (n % 2 == 0 && c != n / 2 || n & 1) break;
        }
      }
    }
    int64_t d1 = id - left_o, d2 = right_o - id, lef = 0;
    if (left_o != -1 && right_o != -1) {
      if (d2 > d1) lef = 1;
    } else if (left_o != -1) {
      lef = 1;
    }
    if (lef) {
      mp[left_o] = 1;
      v[id] = 0;
      ans += d1;
    } else {
      mp[right_o] = 1;
      v[id] = 0;
      ans += d2;
    };
    ;
  }
  cout << ans;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int64_t t = 1;
  if (isQuery) cin >> t;
  while (t--) {
    solve();
    cout << endl;
  }
}
