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;
// #include <ext/pb_ds/assoc_container.hpp>
// #include <chrono> 
// using namespace std::chrono; 
// using namespace __gnu_pbds;

#define ff              first
#define ss              second
#define int             long long
#define double          long double
#define pb              push_back
#define mp              make_pair
#define pi              pair<int,int>
#define vtiii           vector<tuple<int,int,int>>
#define vi              vector<int>
#define vpii            vector<pair<int,int>
#define mii             map<int,int>
#define pqb             priority_queue<int>
#define pqs             priority_queue<int,vi,greater<int>>
#define setbits(x)      __builtin_popcountll(x)
#define zrobits(x)      __builtin_ctzll(x)
#define inf             1e18  
#define ps(x,y)         fixed<<setprecision(y)<<x
#define mk(arr,n,type)  type *arr=new type[n];
#define all(x) (x).begin(), (x).end()
#define FIO             ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0) 
mt19937                 rng(chrono::steady_clock::now().time_since_epoch().count());
 
// typedef tree<int, null_type, less_equal<int>, rb_tree_tag, tree_order_statistics_node_update> indexed_set;

#ifndef ONLINE_JUDGE
#define debug(x) cerr << #x <<" "; _print(x); cerr << endl;
#else
#define debug(x)
#endif

void _print(int t) {cerr << t;}
void _print(string t) {cerr << t;}
void _print(char t) {cerr << t;}
void _print(double t) {cerr << t;}

template <class T, class V> void _print(pair <T, V> p);
template <class T> void _print(vector <T> v);
template <class T> void _print(set <T> v);
template <class T, class V> void _print(map <T, V> v);
template <class T> void _print(multiset <T> v);
template <class T, class V> void _print(pair <T, V> p) {cerr << "{"; _print(p.ff); cerr << ","; _print(p.ss); cerr << "}";}
template <class T> void _print(vector <T> v) {cerr << "[ "; for (T i : v) {_print(i); cerr << " ";} cerr << "]";}
template <class T> void _print(set <T> v) {cerr << "[ "; for (T i : v) {_print(i); cerr << " ";} cerr << "]";}
template <class T> void _print(multiset <T> v) {cerr << "[ "; for (T i : v) {_print(i); cerr << " ";} cerr << "]";}
template <class T, class V> void _print(map <T, V> v) {cerr << "[ "; for (auto i : v) {_print(i); cerr << " ";} cerr << "]";}
 
const int mxN = 2e5 + 10;
const int mod = 1e9 + 7;

void solve(){
    int n;cin>>n;
    vi v(n+2),p(n+2),s(n+2),X,Y;
    X.push_back(0);
    Y.push_back(0);
    for(int i=1;i<=n;++i) {
        cin>>v[i]; p[i] = v[i] + p[i-1];
        if(v[i])X.push_back(i);
        else Y.push_back(i);
    }

    // for(int x : X) cout<<x<<" ";
    //     cout<<endl;
    // for(int x : Y) cout<<x<<" ";
    //     cout<<endl;
    vector<vi> dp(X.size() , vi(Y.size()));
    for(int k = 1 ; k<= X.size()-1; ++k){
        for(int j=k;j<=Y.size()-1;++j){
            dp[k][j] = abs(X[k]-Y[j]) + dp[k-1][j-1];
            if(j!=k) dp[k][j] = min(dp[k][j] , dp[k][j-1]); 
        }
    }
    // for(int i=1;i<=dp.size()-1;++i){
    //     for(int j=1;j<=dp[0].size()-1;++j){
    //         cout<<dp[i][j]<<" ";
    //     }cout<<endl;
    // }
    int ans = inf;
    for(int i=X.size()-1;i<Y.size();++i){
        ans = min(ans , dp[X.size()-1][i]);
    }
    cout<<ans<<endl;
}
                                                                                                            
int32_t main(){
    // auto start = high_resolution_clock::now()
    FIO;
    #ifndef ONLINE_JUDGE
        freopen("input.txt", "r", stdin);
        freopen("output.txt", "w", stdout);
        freopen("Error.txt", "w", stderr);
    #endif
    // precalc();
    // int x; cin>>x; while(x--) 
    solve(); 
    // auto stop = high_resolution_clock::now(); 
    // auto duration = duration_cast<microseconds>(stop - start); 
    // cout << duration.count() << endl;
}