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 = 1e9 + 7;
const long long N = 2e5 + 5;
const long long inf = 1e14;
vector<long long> v1, v2;
long long dp[5001][5001];
long long solvedp(long long i, long long j) {
  if (i > j) return inf;
  if (i < 0) return 0;
  if (dp[i][j] != -1) return dp[i][j];
  if (i == j) {
    return dp[i][j] = abs(v1[i] - v2[j]) + solvedp(i - 1, j - 1);
  }
  dp[i][j] = inf;
  for (long long k = i; k <= min(j, 2 * i); k++) {
    dp[i][j] = min(dp[i][j], solvedp(i - 1, k - 1) + abs(v1[i] - v2[k]));
  }
  return dp[i][j];
}
void solve() {
  memset(dp, -1, sizeof(dp));
  long long n;
  cin >> n;
  for (long long i = 1; i <= n; i++) {
    long long x;
    cin >> x;
    if (x)
      v1.push_back(i);
    else
      v2.push_back(i);
  }
  long long n1 = v1.size(), n2 = v2.size();
  long long ans = solvedp(n1 - 1, n2 - 1);
  cout << ans;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t = 1;
  while (t--) solve();
}
