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 mod2 = 998244353;
const long long inf = 2e18;
const double tiaohe = 0.57721566490153286060651209;
long long oula(long long x) {
  long long res = x;
  for (int i = 2; i <= x / i; ++i) {
    if (x % i == 0) {
      res = res / i * (i - 1);
      while (x % i == 0) x /= i;
    }
  }
  if (x > 1) res = res / x * (x - 1);
  return res;
}
long long quickmod(long long a, long long n, long long m) {
  long long s = 1;
  while (n) {
    if (n & 1) {
      s = s * a % m;
    }
    a = (a * a) % m;
    n = n / 2;
  }
  return s;
}
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
void ex_gcd(long long a, long long b, long long &x, long long &y,
            long long &d) {
  if (!b) {
    d = a, x = 1, y = 0;
  } else {
    ex_gcd(b, a % b, y, x, d);
    y -= x * (a / b);
  }
}
long long inv(long long t, long long p) {
  long long d, x, y;
  ex_gcd(t, p, x, y, d);
  return d == 1 ? (x % p + p) % p : -1;
}
bool isPrime(long long x) {
  if (x == 2) return true;
  if (x % 2 == 0) return false;
  for (long long i = 2; i * i <= x; i++)
    if (x % i == 0) return false;
  return true;
}
inline long long in() {
  char ch = getchar();
  long long x = 0, f = 1;
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
void print(double x) { printf("%.6lf\n", x); }
double eqa = (1 + sqrt(5.0)) / 2.0;
double E = 2.7182818284;
const double pi = acos(-1);
const double eps = 1e-8;
const int N = 2e5 + 50;
int a[N];
int vis[N], p[N];
int tmp[N], tmp2[N];
int vis2[N];
int ddd[N];
int main() {
  int n;
  while (cin >> n) {
    vector<int> v;
    memset(vis, 0, sizeof vis);
    memset(vis2, 0, sizeof vis2);
    int idx = 0;
    for (int i = 1; i <= n; ++i) {
      cin >> a[i];
      if (a[i] == 0)
        v.emplace_back(i);
      else
        p[++idx] = i;
    }
    long long res = 0;
    int sz = v.size() - 1;
    for (int i = 1; i <= idx; ++i) {
      int now = 0, id = 0;
      for (int j = 0; j <= sz; ++j) {
        if (vis2[j]) continue;
        tmp2[++id] = v[j];
        ddd[id] = j;
      }
      for (int j = 1; j <= idx; ++j) {
        if (vis[j]) continue;
        tmp[++now] = j;
      }
      int mi = 2e9, miid = 0, miid2 = 0;
      for (int j = 1; j <= now; ++j) {
        int lpos = lower_bound(tmp2 + 1, tmp2 + 1 + id, p[tmp[j]]) - (tmp2);
        lpos--;
        if (lpos >= 1) {
          int nowcot = abs(p[tmp[j]] - tmp2[lpos]);
          if (nowcot < mi) {
            mi = nowcot;
            miid = j;
            miid2 = lpos;
          }
        }
        if (lpos + 1 <= id) {
          int nowcot = abs(p[tmp[j]] - tmp2[lpos + 1]);
          if (nowcot < mi) {
            mi = nowcot;
            miid = j;
            miid2 = lpos + 1;
          }
        }
      }
      vis[tmp[miid]] = 1;
      vis2[ddd[miid2]] = 1;
      res += mi;
    }
    cout << res << endl;
  }
  return 0;
}
