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 <iostream>
#include <algorithm>
#include <vector>
#include <cmath>
#include <string>
#include <stack>
#include <iomanip>
#include <climits>
#define MOD 1000000007
#define ll long long int
#include <queue>
using namespace std;

vector <bool> is_prime(1000007,true);



void seive()
{
    for (int i=1; i<=1000007; i++)
        {


            {
                if (is_prime[i])
                    for (int num=2*i; num<=1000007; num+=i)
                        {
                            is_prime[num]=false;
                            //if (d[num]==n)
                            // ans=min(ans,num);

                        }
            }



        }


}

ll GCD(ll x,ll y)
{
    ll mx=max(x,y),mn=min(x,y);

    while(mn!=0)
        {
            ll temp=mn;
            mn=mx%mn;
            mx=temp;
        }

    return mx;
}

vector <vector <ll>> vis;
vector <vector <ll>> a;
vector <vector <vector <char>>> dis;
//vector <ll> vis,counttime;


/*void dfs(ll v)
{
    vis[v]=1;

    for (int child:a[v])
        {
            if (vis[child]==0)
                dfs(child);

        }

}
*/
ll dx[4]= {1,0,-1,0};
ll dy[4]= {0,-1,0,1};
char dir[4]= {'D','L','U','R'};
ll n,m;


vector <vector<ll>> bin;
void binary_conv(ll x,ll ind)
{
    vector <ll> a(31,0);
    int k=-1;
    ll xs=x,val=1;
    for (int i=1; i<=x; i*=2)
        {
            k++;
            val=i;

        }
    cout<<val<<" "<<k<<endl;

    for (ll i=val; i>=1; i/=2)
        {
            if (i<=x)
                {
                    x-=i;
                    a[30-k]=1;

                }
            k--;
        }

    bin[ind]=a;

}

ll binpow(ll a,ll b)

{
    if (b==0)
        return 1;
    ll res=binpow(a,b/2);
    if (b%2==0)
        {

            return (res*res)%1000000007;
        }
    else return (a*res*res)%1000000007;

}





ll mod(ll x,ll y)

{
    return max(x-y,y-x);

}
vector <vector <vector <ll>>> dp;

ll grid(ll a[],ll n,ll x, ll k,ll z,bool is_left)
{

    //cout<<x<<endl;
    if (z<0||x<1||x>n)
        {
            //cout<<"break"<<endl;
            return 0;
        }

    if (k==0&&z>=0)
        return a[x-1];





    else
        {

            //  cout<<"dd"<<endl;
//    cout<<dp[x][z][is_left]<<endl;
            if (dp[x][z][is_left]!=-1)
                return dp[x][z][is_left];

            if (is_left)
                {


                    return dp[x][z][is_left]=max(a[x-1]+grid(a,n,x+1,k-1,z,true),a[x-1]+grid(a,n,x-1,k-1,z-1,false));

                }

            else
                {

                    return dp[x][z][is_left]=a[x-1]+grid(a,n,x+1,k-1,z,true);
                }


        }




}





int main()
{
    ll t=1;
//    cin>>t;

    while(t--)
        {
            ll N;
            cin>>N;


            ll a[N];
            vector <ll> x,y;
            for (int i=0; i<N; i++)
                {
                    cin>>a[i];
                    if (a[i]==1)
                        x.push_back(i+1);

                    else y.push_back(i+1);

                }


            ll n=x.size(),m=y.size();

            ll dp[n+1][m+1];


            for (int i=0; i<=n; i++)
                for (int j=0; j<=m; j++)
                    {

                        if (i==0)
                            dp[i][j]=0;

                        else if (j==0)
                            dp[i][j]=MOD;
                    }


            for (int i=1; i<=n; i++)
                {

                    for (int j=1; j<=m; j++)
                        {

                            if (i>j)
                                dp[i][j]=MOD;

                            else
                                {

//                    cout<<mod(x[i-1],y[j-1])<<endl;
                                    dp[i][j]=min(mod(x[i-1],y[j-1])+dp[i-1][j-1],dp[i][j-1]);

                                }

//                            cout<<dp[i][j]<<" ";
                        }

  //                  cout<<endl;
                }
            cout<<dp[n][m]<<endl;

        }









    return 0;
}

