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<iostream> 

using namespace std;

#define maxn 5005

char s[maxn];
int sum_pref[2][maxn];
int sum_map[2 * maxn + 1];
int dp[maxn];

inline int get_lr(int l, int r)
{
	int s0 = sum_pref[0][r] - (l > 0 ? sum_pref[0][l - 1] : 0);
	int s1 = sum_pref[1][r] - (l > 0 ? sum_pref[1][l - 1] : 0);
	if (s0 > s1)
		return s0 - s1;
	return s1 - s0;
}

int main()
{
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	cout.tie(NULL);
	//freopen("input.txt", "r", stdin);
	int t, n;
// 	cin >> t;
    t =1;
	for (int tc = 1; tc <= t; tc++)
	{
		cin >> n ;
		for (int i = 0; i <= 2 * n; i++)
		{
			if (i < n)
			{
				sum_pref[0][i] = sum_pref[1][i] = 0;
				dp[i] = 1e9;
			}
			sum_map[i] = -1;
		}
		for (int i = 0; i < n; i++)
		{
		    int k;
		    cin>>k;
		    s[i] = k+'0';
			sum_pref[s[i] - '0'][i] = i;
			if (i>0)
			{
				sum_pref[0][i] += sum_pref[0][i - 1];
				sum_pref[1][i] += sum_pref[1][i - 1];
			}
		}
		
		int sum = n; // to account for negative values of sum
		sum_map[sum] = 0;
		for (int i = 0; i< n; i++)
		{
			sum += 2 * (s[i] - '0') - 1;
			if (sum_map[sum] != -1)
			{
				int l = sum_map[sum], r = i;
				int val = get_lr(l, r) + (l > 0 ? dp[l - 1] : 0);
				if (dp[r] > val)
					dp[r] = val;
			}
			sum_map[sum] = i + 1;
			if (s[i] == '0')
			{
				if (i == 0)
					dp[i] = 0;
				else if (dp[i] > dp[i - 1])
					dp[i] = dp[i-1];
			}
		}
		cout <<dp[n - 1] << "\n";
	}
}