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 long long mod = 1000000007;
const long long inf = LLONG_MAX;
void IO() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cout.setf(ios::fixed);
  srand(chrono::high_resolution_clock::now().time_since_epoch().count());
}
long long pow(long long x, long long y, long long p = inf) {
  long long res = 1;
  x = x % p;
  if (x == 0) return 0;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
long long invmod(long long a, long long m) { return pow(a, m - 2, m); }
long long cl(long long a, long long x) {
  return a % x == 0 ? a / x : a / x + 1;
}
long long gcd(long long a, long long b) {
  if (a + b == a or a + b == b) return a + b;
  if (a == b) return a;
  if (a > b) return gcd(a - b, b);
  return gcd(a, b - a);
}
long long dp[5000 + 10][5000 + 10];
long long solve(vector<long long> &a, vector<pair<long long, long long> > &pos,
                long long cur, long long last) {
  if (cur == pos.size()) {
    return 0;
  }
  if (a[cur] == 0) return solve(a, pos, cur + 1, last);
  if (pos[cur].first != -1 and pos[cur].first != last) {
    return dp[cur][last] =
               min(abs(cur - pos[cur].first) +
                       solve(a, pos, cur + 1, max(pos[cur].first, last)),
                   abs(cur - pos[cur].second) +
                       solve(a, pos, cur + 1, pos[cur].second));
  }
  if (pos[cur].second != -1 and pos[cur].second != last)
    return dp[cur][last] = abs(cur - pos[cur].second) +
                           solve(a, pos, cur + 1, pos[cur].second);
  return 1000000;
}
void run_time_terror() {
  memset(dp, -1, sizeof(dp));
  long long n;
  cin >> n;
  vector<long long> a(n);
  bool isocc[n];
  memset(isocc, false, sizeof(isocc));
  for (long long i = 0; i < n; ++i) {
    cin >> a[i];
    if (a[i]) {
      isocc[i] = true;
    }
  }
  long long ans = 0;
  vector<pair<long long, long long> > pos(n);
  for (long long i = 0; i < n; ++i) {
    if (a[i] == 0) {
      pos[i] = pos[i - 1];
      continue;
    }
    long long pos1 = -1;
    for (long long j = i - 1; j >= 0; --j) {
      if (isocc[j] == false) {
        if (i > 0 and j != pos[i - 1].first and j > pos[i - 1].first) {
          pos1 = j;
          break;
        } else if (i == 0) {
          pos1 = j;
          break;
        }
      }
    }
    long long pos2 = -1;
    for (long long j = i + 1; j < n; ++j) {
      if (isocc[j] == false) {
        if (i > 0 and j != pos[i - 1].second and j > pos[i - 1].second) {
          pos2 = j;
          break;
        } else if (i == 0) {
          pos2 = j;
          break;
        }
      }
    }
    pos[i].first = pos1;
    pos[i].second = pos2;
  }
  cout << solve(a, pos, 0, -1);
}
int32_t main() {
  cout << setprecision(0);
  IO();
  long long tt = 1;
  for (long long case_no = 1; case_no <= tt; case_no++) {
    run_time_terror();
  }
  return 0;
}
