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;

#define x first
#define y second
typedef long long LL;
typedef pair<int, int> PII;
typedef pair<LL, LL> PLL;
const int N = 5010;
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + 7;
const double eps = 1e-6;
const char nl = '\n';

int n, m;
int a[N];
int f[N][N];
int minv[N], backup[N];

void solve() {
	cin >> n;
	for(int i = 1; i <= n; i++) cin >> a[i];
	vector<int> ones;
	for(int i = 1; i <= n; i++){
		if(a[i] == 1)
			ones.push_back(i);
	}

	int m = ones.size();
	memset(minv, 0x3f, sizeof minv);
	minv[0] = 0;
	for(int i = 1; i <= n; i++){
		if(a[i] == 1) continue;
		memcpy(backup, minv, sizeof minv);
		for(int j = 1; j <= m; j++){
			f[i][j] = backup[j - 1] + abs(i - ones[j - 1]);
			minv[j] = min(minv[j], f[i][j]);
		} 
	}

	int ans = INF;
	for(int i = 1; i <= n; i++){
		if(a[i]) continue;
		ans = min(ans, f[i][m]);
	}
	cout << ans << nl;
}

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);

    int T = 1;
    //cin >> T;
    while(T--) solve();

    return 0;
}
