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.
//Just have Patience + 1...
#include<bits/stdc++.h>
using namespace std;

#define tarzan ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);
#define int long long 

int rr[] = {0, 1, 1, 1, 0, -1, -1, -1};
int cc[] = {1, 1, 0, -1, -1, -1, 0, 1};

const int mod = 1e9 + 7;
const int N = 5050;

int dp[N][N];
vector<int> ones, zeros;

int solve(int n, int m) {
    if(dp[n][m] != -1) return dp[n][m];
    if(n == 0) return 0;
    if(m == 0) return INT_MAX;

    int ans = INT_MAX;
    if(n < m) ans = solve(n, m - 1);
    ans = min(ans, solve(n - 1, m - 1) + abs(ones[n - 1] - zeros[m - 1]));
    dp[n][m] = ans;
    return dp[n][m];
}

signed main() {
    //freopen("input.txt", "r", stdin);
    //freopen("output.txt", "w", stdout);
    tarzan;
    int test = 1;
    //cin >> test;
    while(test--) {
        memset(dp, -1, sizeof dp);
        int n;
        cin >> n;

        int a[n];
        for(int i = 0; i < n; i++) {
            cin >> a[i];

            if(a[i] == 0) zeros.push_back(i);
            else ones.push_back(i);
        }

        int O = ones.size(), Z = zeros.size();
        cout << solve(O, Z) << '\n';
    }
    return 0;
}
