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 ll long long
#define pb push_back
#define task "talltree"
#define pll pair<ll, ll>
#define pii pair<pll, pll>
#define fi first
#define se second
#define ull unsigned long long

using namespace std;
const ll mod = 1e12+7;
const ll N = 5005;

ll n, m, t, ans, k, a[N], b[N], tong,  cnt, q, c[N], st[N], dp[N][N];
vector<pll> adj[N];

string s[N];
ll pw(ll k, ll n)
{
    ll total = 1;
    for(; n; n >>= 1)
    {
        if(n & 1)total = total * k % mod;
        k = k * k % mod;
    }
    return total;
}

void sol()
{
    cin >> n;
    k = 1;
    for(int i = 1; i <= n; i ++)
    {
        cin >> a[i];
        if(a[i])b[++m] = i;
    }
    for(int i = 1; i <= m; i ++)dp[0][i] = mod;
    k = 0;
    for(int i = 1; i <= n; i ++)
    {
        if(a[i] == 0)
        {
            ++k;
            for(int j = m; j > 0; j --)
            {
                dp[i][j] = min(dp[i-1][j-1] + abs(i-b[j]), dp[i-1][j]);
            }
        }
        else for(int j = 1; j <= m; j ++)dp[i][j] = dp[i-1][j];
    }
    cout << dp[n][m];
}
int main()
{
    if(fopen(task".inp", "r")){
       freopen(task".inp", "r", stdin);
       freopen(task".out", "w", stdout);
    }
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int ntest = 1;
    //cin >> ntest;
    while(ntest -- > 0)
    sol();
}
/*
3 5
1 2 3 3 1
4 3 2 5 3
2 6 6 2 5
1 2 2 2
2 1 3 3
3 2 3 3
3 5 1 3
*/
