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>
using namespace std;
using namespace __gnu_pbds;
/******************************************
* AUTHOR : PANDRAVISAM SHIVA SANTHOSH *
* Handle : santhosh_80 *
* INSTITUTION : IIT Palakkad *
******************************************/
#define ll long long int 
#define pii pair<int,int>
#define pll pair<long long, long long>
#define vi vector<int>
#define vll vector<long long >
#define umap unordered_map
#define all(v) (v).begin(),(v).end()
#define pb push_back
#define eb emplace_back
#define f first
#define s second
#define inf LLONG_MAX
#define neg_inf LLONG_MIN
#define next_line cout << "\n"
#define what_is(x) cerr << #x << " is " << x << "\n";
#define rep(i,a,b) for(int i = a; i < b; i++)
#define repj(j,a,b) for(int j = a; j < b; j++)

inline void read0(auto &v, int n) {
	for(int i = 0; i < n; i++){cin >> v[i];}
}
inline void read1(auto &v, int n) {
	for(int i = 1; i <= n; i++){cin >> v[i];}
}
inline void out0(auto &v, int n) {
	for(int i = 0; i < n; i++){cout << v[i] << " ";}
}
inline void out1(auto &v, int n){
	for(int i = 1; i <= n; i++){cout << v[i] << " ";}
}
inline void auto_print(auto &v){
	for(auto x: v)cout << x << " ";
}

const ll mod = 1000000007;
const ll N = 1e6;

int main() {
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	int n; cin >> n;
	vector<int> v(n);
	read0(v, n);

	ll ans = 0;
	vector<int> pos;
	vector<int> occ;

	for(int i = 0; i < n; i++){
		if(v[i] == 0){
			pos.pb(i);
		}
		else{
			occ.pb(i);
		}
	}

	for(int i = 0; i < occ.size(); i++){
		
		int valr = -1, vall = -1;

		int ir = lower_bound(pos.begin(), pos.end(), occ[i]) - pos.begin();

		if(ir != pos.size()){
			valr = pos[ir];
		}
		if(ir != 0){
			vall = pos[ir-1];
		}

		if(vall == -1){
			ans = ans + (valr - occ[i]);
			pos.erase(pos.begin()+ir);
		}
		else if(valr == -1){
			ans = ans + (occ[i] - vall);
			pos.erase(pos.begin()+ir-1);
		}
		else{
			int dist1 = occ[i]-vall;
			int dist2 = valr - occ[i];

			if(dist2 >= dist1){
				ans = ans + dist1;
				pos.erase(pos.begin()+ir-1);
			}
			else{

				if(i == occ.size()-1){
					ans = ans + dist2;
					pos.erase(pos.begin()+ir);
					continue;
				}

				int num_elements = int(pos.size()) - ir;
				if(num_elements >= (int(occ.size())-i)){
					ans = ans + dist2;
					pos.erase(pos.begin()+ir);

				}
				else{
					ans = ans + dist1;
					pos.erase(pos.begin()+ir-1);
				}
			}
		}
	}
	cout << ans << "\n";
 	return 0;

}