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>
#define IOS ios::sync_with_stdio(0),cin.tie(0),cout.tie(0) 
#define X first
#define Y second 
#define nl '\n'
#define AC return 0
#define pb(a) push_back(a) 
#define mst(a,b) memset(a, b, sizeof a)
#define rep(i,n) for(int i = 0; (i)<(n); i++)
#define rep1(i,n) for(int i = 1; (i)<=(n); i++)
#define scd(a) scanf("%lld", &a)
#define scdd(a,b) scanf("%lld%lld", &a, &b)
#define scs(s) scanf("%s", s)
//#pragma GCC optimize(2)
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair <int, int> pii;
typedef pair <ll, ll> pll;
const ll INF = (ll)0x3f3f3f3f3f3f3f, MAX = 9e18, MIN = -9e18;
const int N = 5e3+10, M = 2e6+10, mod = 1e9+7, inf = 0x3f3f3f3f;
ll a[N], dp[N][N]; 
vector<ll> p1, p2;

int main()
{
	IOS;
	int _;
	// cin>>_;
	_ = 1;
	while(_--)
	{
		ll ans = 0, n;
		cin>>n;
		rep1(i, n)
		{
			cin>>a[i];
			if(a[i])
				p1.pb(i);
			else
			 	p2.pb(i);
		}
		rep1(i, p1.size())
			rep(j, p2.size()+1)
				dp[i][j] = MAX;
		//这里dp[0][i] = 0
		//dp[i][j]表示前i个1与前j个0匹配的最优解
		rep1(i, p1.size())
			rep1(j, p2.size())
			{
				dp[i][j] = dp[i][j-1];
				dp[i][j] = min(dp[i][j], dp[i-1][j-1]+abs(p1[i-1]-p2[j-1]));	
			}
		cout<<dp[p1.size()][p2.size()]<<nl;
	}
	
	
	AC;
} 
