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 int maxn = 5010;
const int maxc = 26;
const int base = 131;
const long long inf = 0x3f3f3f3f3f3f3f3f;
int sum0[maxn], sum1[maxn];
long long dp[maxn][maxn];
int a[maxn];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) cin >> a[i];
  a[0] = 1;
  a[n + 1] = 1;
  for (int i = 1; i <= n; i++) {
    sum0[i] += sum0[i - 1] + (a[i] == 0);
    sum1[i] += sum1[i - 1] + a[i];
  }
  memset(dp, inf, sizeof(dp));
  for (int i = 1; i <= n; i++) {
    if (i < n && (a[i] ^ a[i + 1])) dp[i][i + 1] = 1;
    if (i < n && !a[i] && !a[i + 1]) dp[i][i + 1] = 0;
    if (!a[i]) dp[i][i] = 0;
  }
  for (int len = 3; len <= n; len++) {
    for (int i = 1; i <= n; i++) {
      int j = i + len - 1;
      if (j > n) break;
      if (a[j] ^ a[j - 1]) dp[i][j] = min(dp[i][j], dp[i][j - 2] + 1);
      if (a[i] ^ a[i + 1]) dp[i][j] = min(dp[i][j], dp[i + 2][j] + 1);
      if (a[i] ^ a[j]) dp[i][j] = min(dp[i + 1][j - 1] + j - i, dp[i][j]);
      if (!a[j]) dp[i][j] = min(dp[i][j - 1], dp[i][j]);
      if (!a[i]) dp[i][j] = min(dp[i + 1][j], dp[i][j]);
      for (int k = j; k < j; k++)
        dp[i][j] = min(dp[i][j], dp[i][k] + dp[k + 1][j]);
    }
  }
  cout << dp[1][n];
  return 0;
}
