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 pb push_back
#define pii pair<int,int>
#define pll pair<ll, ll>
#define pil pair<int,ll>
#define pli pair<ll,int>
#define ppi pair<pii,int>
#define pip pair<int,pii>
#define ppp pair<pii,pii>
#define pdd pair<double, double>
#define f first
#define s second
#define MOD 1000000007
#define mkp make_pair
#define FOR(i,l,r) for (int i=l;i<=r;i++)
#define LOR(i,l,r) for (ll i=l;i<=r;i++)
#define FORD(i,r,l) for (int i=r;i>=l;i--)
#define LORD(i,r,l) for (ll i=r;i>=l;i--)
#define CL(x) memset(x,0,sizeof(x))
#define ALL(x) x.begin(), x.end()
#define SZ(x) (int)(x.size())
#define UI(x) (int)(x-'A')
#define LI(x) (int)(x-'a')
#define DI(x) (int)(x-'0')
#define LL_MAX (1LL<<60)
#define DB 0
#define DBG(x) if(DB)cout << #x << " : " << x << '\n'
#define PRALL(v) if(DB)for(auto it : v)cout<<v<<' '; NL;
#define NL cout<<'\n';
#define FND(S,x) (S.find(x)!=S.end())
#define bit(x,y) (!!(((ll)x)&(1LL<<y)))
typedef long long ll;
typedef long double ld;

#define MXN 5001
int N;
int n, m;
vector<int> O, V;
ll DP[MXN][MXN];

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);

    cin>>N;
    int ai;
    FOR(i,1,N)
    {
        cin>>ai;
        if(ai==0)V.pb(i);
        else O.pb(i);
    }
    n = SZ(O);
    m = SZ(V);
    FOR(i,0,n)FOR(j,0,m)DP[i][j]=LL_MAX;
    FOR(i,0,m)DP[0][i] = 0;
    FOR(i,1,n)
    {
        FOR(j,i,m)
        {
            DP[i][j] = min( DP[i][j-1], DP[i-1][j-1]+abs(V[j-1]-O[i-1]) );
        }
    }
    //cout<<DP[1][2]<<'\n';
    cout<<DP[n][m]<<'\n';

    return 0;
}
