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 <vector>
#include <unordered_map>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <set>
#include <map>
#include <queue>

#define ll long long int
#define mod 1000000007

using namespace std;

void input(vector<ll> &arr)
{
    for(int i=0;i<arr.size();i++)
      cin>>arr[i];
}

void output(vector<ll> arr)
{
   for(int i=0;i<arr.size();i++)
     cout<<arr[i]<<" ";

   cout<<"\n";
}

ll gcd(ll a,ll b)
{
   if(b==0)
     return a;

   return gcd(b,a%b);
}

ll power(ll a,ll b)
{
    if(b==0)
      return 1;

    if(b==1)
      return a;

    ll smallans=power(a,b/2);
    ll myans=(smallans*smallans)%mod;

    if((b&1))
      myans=(myans*a)%mod;

    return myans;
}

ll multiply(ll a,ll b)
{
   ll ans=((a%mod)*(b%mod))%mod;

   return ans;
}

ll add(ll a,ll b)
{
   ll ans= ((a%mod)+(b%mod))%mod;

   return ans;
}

ll subtract(ll a,ll b)
{
   ll ans=((a%mod)-(b%mod))%mod;

   if(ans<0)
    ans+=mod;

   return ans;
}

ll divide(ll a,ll b)
{
   return multiply(a,power(b,mod-2));
}

//======================== HASH FUNCTION FOR UNORDERED MAP OF PAIRS =============================

struct hash_pair { 
    template <class T1, class T2> 
    size_t operator()(const pair<T1, T2>& p) const
    { 
        auto hash1 = hash<T1>{}(p.first); 
        auto hash2 = hash<T2>{}(p.second); 
        return hash1 ^ hash2; 
    } 
}; 

//========================== BINARY INDEX TREE ===================================================

void update(vector<ll> &tree,ll index,ll val)
{
   index++;
   while(index < tree.size())
   {
      tree[index]=(tree[index] + val) %mod;
 
      index+=index&(-index);
   }
}
 
ll calculate(vector<ll> &tree,ll index)
{
   ll sum=0;
   index++;
   while(index > 0)
   {
      sum=(sum + tree[index])%mod;
 
      index-=index&(-index);
   }
 
   return sum;
}

//=========================== FOR DISJOINT SET UNION =========================================

ll findpar(ll p,vector<ll> &parent)
{
    if(parent[p]==p)
     return p;

    parent[p]=findpar(parent[p],parent);

    return parent[p];
}

void merge(ll a, ll b,vector<ll> &parent,vector<ll> &size_) 
{
    a = findpar(a,parent);
    b = findpar(b,parent);
    
    if (a != b) 
    {
        if (size_[a] < size_[b])
             swap(a, b);
             
        parent[b] = a;
        size_[a] += size_[b];
    }
}

//   ====================================== FOR STORING AND COUNTING THE PRIMES USING SIEVE ==============

void sieve(vector<bool> &primes,vector<ll> &count)
{
   for(int i=2;i<primes.size();i++)
   {
      if(primes[i])
      {
         count.push_back(i);
         for(int j=i*i;j<primes.size();j+=i)
           primes[j]=false;
      }
   }
}

// ==================================================================================================


void solve()
{
    ll n;
    cin >> n;

    vector <ll> arr(n,0);
    input(arr);

    
    map<ll,ll> mp1;
    map<ll,ll> mp2;

    ll one = 0;
    ll zero = 0;

    for(int i = 0;i < n;i++)
    {
       if(arr[i] == 1)
       {
          mp2[one] = i;
          one ++;
       }

       else
       {
          mp1[zero] = i;
          zero ++;
       }
    }

    if(one == 0)
    {
       cout << 0 <<"\n";
       return;
    }

    vector < vector < ll> > dp(one,vector <ll>(zero,1e18));
    ll maxi = 1e18;

    for(int i = 0;i < one;i++)
    {
       for(int j = 0;j < zero;j++)
       {
          if(i == 0)
          {
             if(j == 0)
             dp[i][j] = min(maxi,abs(mp1[j] - mp2[i]));

             else dp[i][j] = min(dp[i][j - 1],abs(mp1[j] - mp2[i]));
          }

          else
          {
             if(j < i)
               continue;

             dp[i][j] = min(dp[i][j - 1],dp[i - 1][j - 1] + abs(mp1[j] - mp2[i]));

          }

         
       }
    }
    
    cout << dp[one - 1][zero - 1] << "\n";

   
    
}

int main()
{
   ios_base::sync_with_stdio(false); 
   cin.tie(NULL);
   
   int t = 1;
   // cin>>t;

   while(t--)
   {

      solve();
      
   }

   return 0;
}