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 N = 5002, inf = N * N * N;
long long n;
long long dp[N][N];
vector<long long> a[2];
long long solve(long long i, long long j) {
  if (i == (long long)a[1].size()) {
    return 0;
  }
  if (j == (long long)a[0].size()) {
    if (i != (long long)a[1].size()) {
      return inf;
    } else {
      return 0;
    }
  }
  long long &ans = dp[i][j];
  if (ans != -1) return ans;
  ans = inf;
  ans = min(solve(i + 1, j + 1) + abs(a[1][i] - a[0][j]), solve(i, j + 1));
  return ans;
}
int32_t main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> n;
  for (long long i = 0; i < n; ++i) {
    long long ai;
    cin >> ai;
    a[ai].push_back(i);
  }
  if (a[1].empty()) {
    return cout << 0, 0;
  }
  auto rem = [&](long long sz, long long cur) { return sz - cur; };
  memset(dp, 2, sizeof dp);
  dp[0][0] = 0;
  for (long long i = 0; i < (long long)a[1].size(); ++i) {
    for (long long j = 0; j < (long long)a[0].size(); ++j) {
      if (rem(a[0].size(), j) >= rem(a[1].size(), i)) {
        dp[i + 1][j + 1] =
            min(dp[i + 1][j + 1], dp[i][j] + abs(a[1][i] - a[0][j]));
      }
      if (rem(a[0].size(), j + 1) >= rem(a[1].size(), i)) {
        dp[i][j + 1] = min(dp[i][j + 1], dp[i][j]);
      }
    }
  }
  cout << dp[a[1].size()][a[0].size()] << '\n';
  return 0;
}
