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>
#define ff first
#define ss second
#define pb push_back
#define pf push_front
#define ins insert
#define len(s) s.length()
#define forp(i, a, b) for (int i = a; i <= b; i++)
#define rep(i, n) for (int i = 0; i < n; i++)
#define ren(i, n) for (int i = n - 1; i >= 0; i--)
#define forn(i, a, b) for (int i = a; i >= b; i--)
#define on cout << endl
#define all(v) v.begin(), v.end()
#define mem(n, m) memset(n, m, sizeof(n))
#define pii pair<int, int>
#define pll pair<long long, long long>
#define sii set<int>
#define us unordered_set<int>
#define um unordered_map<int, int>
#define sll set<long long>
#define vii vector<int>
#define vll vector<long long>
#define mll map<long long, long long>
#define lob lower_bound
#define upb upper_bound
#define ret return 0

#define boost ios_base::sync_with_stdio(0)
#define MOD 1000000007
#define EPSILON 1e-9
#define PI 3.14159265358979323846
#define inf 999999999999999999
#define siz 100001
#define SIZ 1000001
#define SIZE 200001

typedef long long ll;
typedef long double ldo;
typedef double db;
using namespace std;
#define pqb priority_queue<int>
#define pqs priority_queue<int, vi, greater<int>>
#define setbits(x) __builtin_popcountll(x)
#define zrobits(x) __builtin_ctzll(x)
#define w(tt)  \
    int tt;    \
    cin >> tt; \
    while (tt--)

void c_p_c()
{
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
#ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
#endif
}
const int N = 2e5 + 7;
int n;
int m1, m2;
vector<vector<int>> dp;
vector<int> zero, one;
int sol(int i, int j)
{
    if (j == m2)//zero metrix khtm
    {
        return 0;
    }
    if (i == m1)//one matrix khtm
    {
        return 1e9;
    }
    if (dp[i][j] != -1)
    {
        return dp[i][j];
    }

    return dp[i][j] = min(sol(i + 1, j), sol(i + 1, j + 1) + abs(one[j] - zero[i]));
}
int main()
{
    c_p_c();
    cin >> n;
    for (int i = 0; i < n; i++)
    {
        int ele;
        cin >> ele;
        if (ele == 1)
        {
            one.push_back(i);
        }
        else
        {
            zero.push_back(i);
        }
    }
    m1 = zero.size(), m2 = one.size();
    dp = vector<vector<int>>(m1, vector<int>(m2, -1));
    cout << sol(0, 0);

    return 0;
}
// author -> @__Aditya__