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;
typedef long long ll;
#define endl "\n"
#define quick ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL)
#define ALL(x) x.begin(),x.end()
#define SORT(x) sort(ALL(x))
#define FOR(a,b,c) for(ll (a)= (b);(a)<(c);++(a))
#define REP(a,b) FOR(a,0,b)
#define mp make_pair
#define pb push_back
#define pll pair<ll,ll>
#define f first
#define s second
ll t = 1;

template <class myType>
void print_arr(myType &arr, ll L, string sep){
    REP(i,L){
        cout << arr[i];
        if (i < L-1){
            cout << sep;
        }
    }
    cout << endl;
    return;
}

/*

*/

void solve(){
    int n;
    cin >> n;
    vector<int> a(n+1);
    REP(i,n) cin >> a[i+1];
    vector<int> pos;
    for(int i = 1; i <= n; i++) {
        if (a[i] == 1) pos.pb(i);
    }
    vector<vector<int>> dp(pos.size()+1, vector<int> (n+1,1e9));
    REP(i,n+1) dp[0][i] = 0;
    for (int i = 1; i <= pos.size(); i++) {
        for (int j = 1; j <= n; j++) {
            dp[i][j] = min(dp[i][j],dp[i][j-1]);
            if (a[j] == 1) continue;
            dp[i][j] = min(dp[i][j],dp[i-1][j-1] + abs(pos[i-1]-j));
            //cout << i << " " << j << " " << dp[i][j] << endl;
        }
    }
    int ans = 1e9;
    REP(i,n+1) {
        ans = min(ans,dp[pos.size()][i]);
    }
    cout << ans << endl;
    return;
}


int main(){  
    quick;
    //cin >> t;
    while (t--) solve();
    return 0;
}
