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.
#pragma GCC optimize("-Ofast","unroll-loops","omit-frame-pointer","inline") //Optimization flags
#pragma GCC option("arch=native","tune=native","no-zero-upper") //Enable AVX
#pragma GCC target("avx2")  //Enable AVX
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>

using namespace __gnu_pbds;
using namespace std;
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
typedef tree<int,null_type,less<int>,rb_tree_tag,tree_order_statistics_node_update> indexed_set;
//order_of_key(x) : number of items less than x
//find_by_order(x) : iterator to x

#define mem(dp,a)           memset(dp,a,sizeof dp)
#define rep(i,a,b)          for(ll i=a;i<b;i++)
#define pb(x)               push_back(x)
#define mp(x,y)             make_pair(x,y)
#define fastio              ios_base::sync_with_stdio(false);cin.tie(NULL)
#define F                   first
#define S                   second
#define all(v)              (v).begin(),(v).end()

const ld pi=3.14159265359;
ll INF=1e18+10;
ll MOD=998244353;
ll mod=1e9+7;

inline ll add(ll a,ll b,ll m){
    if((a+b)>=m)return (a+b)%m;
    return a+b;
}

inline ll mul(ll a,ll b,ll m){
    if((a*b)<m)return a*b;
    return (a*b)%m;
}

int main(){
    // freopen("in.txt", "r", stdin);
    // freopen("out.txt", "w", stdout);

    int n;cin>>n;
    set<int> z, o;
    rep(i,0,n){
        int a;cin>>a;
        if(a == 0)z.insert(i);
        else o.insert(i);
    }

    ll ans = 0;
    for(auto x: o){
        auto it = z.lower_bound(x);
        if(it == z.begin()){
            int itp = *it;
            ans += abs(x-itp);
            z.erase(it);
        } else if(it == z.end()){
            it--;
            int itp = *it;
            ans += abs(x-itp);
            z.erase(it);
        }
        else {
            int itp1 = *it;
            it--;
            int itp2 = *it;
            if(abs(itp2-x) <= abs(itp1-x)){
                ans += abs(itp2-x);
                z.erase(itp2);
            } else {
                ans += abs(itp1-x);
                z.erase(itp1);
            }
        }
    }

    cout<<ans<<endl;
}