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 ll long long
const long long MOD = 1e9 + 7;
const long long INF = 1e9 + 7;
#define pb push_back
#define ff first
#define ss second
#define max_(a,b,c) max1(max1(a,b),c)
#define min_(a,b,c) mn(mn(a,b),c)
long gcd(long a,long b) { return b?gcd(b,a%b):a;}
#define rep(i,a,n) for(int i=a;i<n;i++)
#define per(i,n,a) for(int i=n-1;i>=a;i--)
#define vi vector<int>
#define pii pair<int,int>
#define pll pair<long long,long long>
#define vll vector<long long>
const int mod = 1e9 +7;
const int inf=INT_MAX;
const ll inff=LLONG_MAX;
int mem[5005][5005];
int dp(vector<int> &a, vector<int> &b, int i, int j)
{
    if(mem[i][j]!=-1)
        return mem[i][j];

    if(i>=a.size())
        return 0;

    if(j>=b.size())
        return inf;
    
    int temp=inf;
    int paired=dp(a,b,i+1,j+1);
    int not_paired=dp(a,b,i,j+1);
    if(paired!=inf)
    {
        temp=min(temp,abs(a[i]-b[j])+paired);
    }
    if(not_paired!=inf)
    {
        temp=min(temp,not_paired);
    }
    mem[i][j]=temp;
    return temp;
}
void process()
{
    int n;
    cin>>n;
    vector<int> v;
    vector<int> a1,a0;
    rep(i,0,n)
    {
        int x;
        cin>>x;
        if(x==1)
        {
            a1.push_back(i);
        }
        else if(x==0)
        {
            a0.push_back(i);
        }
        v.push_back(x);
    }
    for(int i=0;i<n+2;++i)
    {
        for(int j=0;j<n+2;++j)
        {
            mem[i][j] = -1;
        }
    }
    cout<<dp(a1,a0,0,0)<<"\n";
}

int main()
{
    ios::sync_with_stdio(0);
    cin.tie(0);
    #ifndef ONLINE_JUDGE
    //cout<<"Hi!";
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
    //  chrono::time_point<std::chrono::system_clock> start, end;
    //  start = chrono::system_clock::now(); 
    #endif // ONLINE_JUDGE
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    ll t=1;
    //cin>>t;
    while(t--)
        process();
    return 0;
}