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>

/**
******************* Author:Bisnu sarkar ****************************
**/
/*
#pragma GCC target ("avx2")
#pragma GCC optimization ("O3")
#pragma GCC optimization ("unroll-loops")
*/

//PBDS
/*
#include<ext/pb_ds/assoc_container.hpp>
#include<ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
*/

using namespace std;

//PBDS_SET
//Index of Value - .order_of_key()
//Value at Index - .find_by_order()
/*
typedef tree<int, null_type, less<int>, rb_tree_tag,tree_order_statistics_node_update> pbds_set; 
typedef tree<int, null_type, less_equal<int>, rb_tree_tag,tree_order_statistics_node_update> pbds_multiset; 
*/

#define fRead(x)       freopen(x,"r",stdin)
#define fWrite(x)      freopen (x,"w",stdout)
#define ull            unsigned long long
#define ll             long long
#define pii            pair<int,int>
#define sit            set<int> :: iterator
#define vrit           vector<int> :: reverse iterator
#define ff             first
#define ss             second
#define endl           '\n';
#define sz(s)          (int)s.size()
#define all(s)         s.begin(),s.end()
#define IO             ios_base::sync_with_stdio(false),cin.tie(NULL);  
#define what_is(x)     cerr << #x << " is " << x << endl;
///*....Debugger....*///
#define error(args...) { string _s = #args; replace(_s.begin(), _s.end(), ',', ' '); stringstream _ss(_s); istream_iterator<string> _it(_ss); err(_it, args); }
void err(istream_iterator<string> it) {cout << endl;}
template<typename T, typename... Args> void err(istream_iterator<string> it, T a, Args... args) {
  cerr << *it << " = " << a << ", ";
  err(++it, args...);
}
///*....Input....*///
template <typename T> inline void Int(T &n) {
  n = 0; int f = 1; register int ch = getchar();
  for (; !isdigit(ch); ch = getchar()) if (ch == '-') f = -1;
  for (; isdigit(ch); ch = getchar()) n = (n << 3) + (n << 1) + ch - '0';
  n = n * f;
}
 
template <typename T, typename TT> inline void Int(T &n, TT &m) { Int(n); Int(m); }
template <typename T, typename TT, typename TTT> inline void Int(T &n, TT &m, TTT &l) { Int(n, m); Int(l);}
double ind(){double x; scanf("%lf",&x);return x;}

///*....Bitmask....*///
int set_1(int n,int pos){return n = (n | (1<<pos));}
int reset_0(int n,int pos){return n= n & ~(1<<pos);}
bool check_bit(int n,int pos){return n = n & (1<<pos);}

int dx[8]= {1,0,-1,0,-1,-1,1,1};
int dy[8]= {0,1,0,-1,-1,1,-1,1};

///*....Constraints....*///
const int N = (int) 5e3 + 5;
const int M = (int) 1e9 + 7;
const double pi=2* acos(0.0);
const double eps=1e-9;

int a[N],dp[N][N],m,n;
std::vector<int> v;
int fun(int pos,int ones)
{
  if(ones >= m)return 0;
  if(pos >=n)return M;

  int &ret = dp[pos][ones];
  if(ret != -1)return ret;

  ret=fun(pos+1,ones);

  if(a[pos+1] == 0){
    ret = min(ret,abs(pos+1-v[ones])+fun(pos+1,ones+1));
  } 
  return ret;
}

int main()
{
    int t=1,te=0;
    //Int(t);
    while(t--){
        
        Int(n);

        for(int i=1;i<=n;++i){
          Int(a[i]);
          if(a[i] == 1)v.push_back(i);
        }          
        m=sz(v);
        memset(dp,-1,sizeof(dp));

        printf("%d\n", fun(0,0));

        //printf("Case %d: ",++te);
    }

    //cerr << "Time elapsed :" << clock() * 1000.0 / CLOCKS_PER_SEC << " ms" << '\n';
    return 0;
}




