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;
#define int long long
#define vi vector<int>
#define vpii vector<pair<int,int>>
#define pii pair<int,int>
#define ff first
#define ss second
#define print(x) cout<<x<<endl
#define printt(x,y) cout<<x<<" "<<y<<"\n"
#define pb push_back
#define all(v) v.begin(),v.end()
#define sz(x) (int)x.size()
#define vvi vector<vector<int>>
#define maxq priority_queue<int>
#define minq priority_queue<int,vector<int>,greater<int>>
const int mod = 1e9 + 7;
const int N = 200005;

int dp[5005][5005];
vi a, b;
int sol(int i, int j) {

    if (i == sz(a))
        return 0;

    if (j == sz(b))
        return 1e9;

    if (dp[i][j] != -1)
        return dp[i][j];

    int ans = 1e9;

    ans = min(ans, abs(a[i] - b[j]) + sol(i + 1, j + 1));

    ans = min(ans, sol(i, j + 1));

    return dp[i][j] = ans;
}

void solve() {

    int n;
    cin >> n;

    for (int i = 0; i < n; i++) {
        int u;
        cin >> u;
        if (u == 1)
            a.pb(i);
        else
            b.pb(i);
    }
    memset(dp, -1, sizeof(dp));
    int ans = sol(0, 0);

    print(ans);

}

int32_t main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
#ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
#endif
    int t = 1;
    //cin >> t;
    while (t--) {
        solve();
    }
}