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;
void solve() {
  long long int a, b, c, d, m, n, p, q, r, x, y, z, aa, bb, cc, dd;
  double pp, qq, rr, xx, yy, zz;
  char ch1, ch2, ch3, ch4, ch5;
  long long int i, j, k, l;
  long long int cnt = 0, cnt1 = 0, cnt2 = 0, cnt3 = 0, sum = 0, sum2 = 0,
                mx = LLONG_MIN, mn = LLONG_MAX;
  long long int flag = 0, flag2 = 0;
  vector<long long int> vec1, vec2, vec3;
  string s1, s2, s3;
  cin >> n;
  for (long long int i = (long long int)(0); i < (long long int)(n); ++i) {
    cin >> x;
    vec1.push_back(x);
  }
  vec2 = vec1;
  for (long long int i = (long long int)(0); i < (long long int)(n); ++i) {
    if (vec1[i] == 1) {
      j = i;
      k = i;
      flag = 0;
      flag2 = 0;
      while (j >= 0) {
        if (vec1[j] == 0) {
          flag = 1;
          x = j;
          break;
        }
        j--;
      }
      while (k < n) {
        if (vec1[k] == 0) {
          flag2 = 1;
          y = k;
          break;
        }
        k++;
      }
      if (flag && flag2) {
        a = i - x;
        b = y - i;
        if (a <= b) {
          vec1[x] = 2;
          sum += a;
        } else {
          vec1[y] = 2;
          sum += b;
        }
      } else if (flag) {
        a = i - x;
        vec1[x] = 2;
        sum += a;
      } else {
        b = y - i;
        vec1[y] = 2;
        sum += b;
      }
    }
  }
  vec1 = vec2;
  for (long long int i = (long long int)(n - 1); i >= (long long int)(0); --i) {
    if (vec1[i] == 1) {
      j = i;
      k = i;
      flag = 0;
      flag2 = 0;
      while (j >= 0) {
        if (vec1[j] == 0) {
          flag = 1;
          x = j;
          break;
        }
        j--;
      }
      while (k < n) {
        if (vec1[k] == 0) {
          flag2 = 1;
          y = k;
          break;
        }
        k++;
      }
      if (flag && flag2) {
        a = i - x;
        b = y - i;
        if (a < b) {
          vec1[x] = 2;
          sum2 += a;
        } else {
          vec1[y] = 2;
          sum2 += b;
        }
      } else if (flag) {
        a = i - x;
        vec1[x] = 2;
        sum2 += a;
      } else {
        b = y - i;
        vec1[y] = 2;
        sum2 += b;
      }
    }
  }
  cout << min(sum, sum2) << endl;
}
int main() {
  ios::sync_with_stdio(false);
  long long int t = 1;
  while (t--) {
    solve();
  }
  return 0;
}
