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>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

using namespace __gnu_pbds;
using namespace std;
using   ll = long long;
using   ld = long double;

#define fast            ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define ordered_set     tree<int, null_type,less<int>, rb_tree_tag,tree_order_statistics_node_update>

#define pb              push_back
#define mp              make_pair
#define F               first
#define S               second
#define pii             pair<int , int>
#define int             long long int
#define endl            "\n"


#define ALL(v)          v.begin(),v.end()
#define ALLR(v)         v.rbegin(),v.rend()
#define sz(v)           (int) v.size()
#define PI              3.14159265358979323
#define inf             LLONG_MAX
#define ones(x)         __builtin_popcount(x)
#define mod             1000000007
#define MOD             998244353

ll mod_pow(ll a,ll b,ll m)
{
	ll res = 1;
	while(b)
	{
		if(b&1)
		{
			res=(res*a) % m;
		}
		a=(a*a) % m;
		b>>=1;
	}
	return res;
}

ll mod_inverse(int a , int m)
{
	return mod_pow(a , m - 2 , m);
}

const int N = 5005;

int dp[N][2];
int n;
vector< int > pos;
set< int > st;

int calc(int i , int op)
{
        if(i >= pos.size())
        {
                return 0;
        }
        
        if(dp[i][op] != -1)
        {
                return dp[i][op];
        }
        
        if(op == 1)
        {
                int ans = inf;
                auto u = st.upper_bound(pos[i]);
        
                if(u != st.end())
                {
                        int j = *u;
                        st.erase(u);
                        ans = min(ans , abs(pos[i] - j) + min(calc(i + 1 , 1) , calc(i + 1 , 0)));
                        st.insert(j);
                }
                
                return dp[i][op] = ans;
        }
        else
        {
                int ans = inf;
                
                auto v = st.upper_bound(pos[i]);
        
                if(v != st.begin())
                {
                        v--;
                        int j = *v;
                        st.erase(v);
                        ans = min(ans , abs(pos[i] - j) + min(calc(i + 1 , 1) , calc(i + 1 , 0)));
                        st.insert(j);
                }

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

void solve()
{
      cin >> n;
      
      int a[n];
      
      for(int i = 0; i < n; ++i)
      {
              dp[i][0] = dp[i][1] = -1;
      }
      
      for(int i = 0; i < n; ++i)
      {
              cin >> a[i];
              
              if(a[i] == 1)
              {
                      pos.pb(i);
              }
              else
              {
                      st.insert(i);
              }
      }
      
      cout << min(calc(0 , 0) , calc(0 , 1)) << endl;
}

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

	while(t--) {
		solve();
	}

	return 0;
}