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;

typedef long long ll;
typedef long double lld;
typedef vector<ll> vll;
typedef vector<vll> vvll;
typedef pair<ll, ll> pll;
typedef vector<pll> vpll;
typedef vector<bool> vb;

#define fi first
#define se second
#define mp make_pair
#define pb push_back
#define len(a) a.size()
#define trace(...)
#define newl '\n'
#define gcd __gcd
#define forl(i, k, n) for (ll i = k; i < n; i++)
#define fore(i, k, n) for (ll i = k; i <= n; i++)
#define revg(i, k, n) for (ll i = k; i > n; i--)
#define reve(i, k, n) for (ll i = k; i >= n; i--)
#define MOD 1000000007 //10^9 + 7
#define all(x) x.begin(), x.end()
#define len(a) a.size()
#define flush fflush(stdout)
#define rz(n,k) resize(n,k)
#define printG(g) forl(i,0,g.size()){cout<<i<<"->";for(ll x:g[i]) cout<<x<<' ';cout<<newl;}
#define print(x)                                   \
    for (auto it = x.begin(); it != x.end(); it++) \
        cout << *it << ' ';                        \
    cout << newl;
#define fastio                        \
    ios_base::sync_with_stdio(false); \
    cin.tie(NULL);

ll N = 1e5 + 1, LOGN = 25;
ll inf = 1e8 + 1;
// vll vis(N, 0);

void sudhirMTS()
{
    ll n;
    cin >> n;
    vll a(n, 0);
    vll o, z;
    forl(i, 0, n) {
        cin >> a[i];
        if (a[i] == 0) z.pb(i);
        else o.pb(i);
    }
    vvll dp(z.size() + 1, vll (o.size() + 1, inf));
    dp[0][0] = 0;
    forl(i, 1, z.size() + 1) {
        dp[i][0] = 0;
        forl(j, 1, o.size() + 1) {
            dp[i][j] = min(dp[i - 1][j - 1] + abs(z[i - 1] - o[j - 1]), dp[i - 1][j]);
        }
    }
    cout << dp[z.size()][o.size()] << newl;
}

int main()
{
    fastio;
#ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
    freopen("error.txt", "w", stderr);
    freopen("output.txt", "w", stdout);
#endif
    int t = 1;
    // cin >> t;
    while (t--)
    {
        sudhirMTS();
    }
    cerr << "time taken : " << (float)clock() / CLOCKS_PER_SEC << " secs" << endl;
    return 0;
}