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.
// MD. Ashiqur Rahman
#include<bits/stdc++.h>
using namespace std;

#define         ll         long long
#define         pb         push_back
#define         scn(x)     scanf("%d",&x)
#define         scnl(x)    scanf("%lld",&x)
#define         prnt(x)    printf("%d\n",x)
#define         prntl(x)   printf("%lld\n",x)
#define         prii       pair<int,int>
#define         mapii      map<int,int>
#define         mapll      map<ll,ll>
#define         mapci      map<char,int>
#define         mapcl      map<char,ll>
#define         mapsi      map<string,int>
#define         mapsl      map<string,ll>
#define         prll       pair<ll,ll>
#define         vctri      vector<int>
#define         vctrl      vector<ll>
#define         vctrd      vector<double,double>
#define         all(a)     (a).begin(),(a).end()
#define         rall(a)    (a).rbegin(),(a).rend()
#define         F          first
#define         S          second
#define         mp         make_pair
#define         ftc(x)     cerr << #x << ": " << x << " " << '\n';
#define         PI         acos(-1)
#define         lcm(a,b)   ((a*b)/__gcd(a,b))
#define         fast       ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define         sqr(a)     ((a)*(a))
#define         memset(x,v) memset(x, v, sizeof(x))
#define         ask         '\n'
#define         negmod(x,y) ((x % y) + y) % y 


ll MOD;
inline void modulas(ll a) {MOD = a;}
inline void normal(ll &a) { a %= MOD; (a < 0) && (a += MOD); }
inline ll modMul(ll a, ll b) { a %= MOD, b %= MOD; normal(a), normal(b); return (a*b)%MOD; }
inline ll modAdd(ll a, ll b) { a %= MOD, b %= MOD; normal(a), normal(b); return (a+b)%MOD; }
inline ll modSub(ll a, ll b) { a %= MOD, b %= MOD; normal(a), normal(b); a -= b; normal(a); return a; }
inline ll modPow(ll b, ll p) { ll r = 1; while(p) { if(p&1) r = modMul(r, b); b = modMul(b, b); p >>= 1; } return r; }
inline ll modInverse(ll a) { return modPow(a, MOD-2); }
inline ll modDiv(ll a, ll b) { return modMul(a, modInverse(b)); }


vector<bool> isPrime(10000010, true);
inline void seivePrime(ll L, ll R) { ll lim = sqrt(R);for (ll i = 2; i <= lim; ++i){
        for (ll j = max(i * i, (L + i - 1) / i * i); j <= R; j += i)
            isPrime[j - L] = false;}if (L == 1)isPrime[0] = false;}
inline ll chckPrime(ll L,ll prme){return isPrime[prme - L];}
inline ll cntPrime(ll L,ll R){return count(isPrime.begin(),isPrime.begin() + R - L + 1,true);}



int main(){
    fast;
    int t = 1;
    // cin >> t;
    while(t --){
        int n;
        cin >> n;
        int ar[n];
        for(int i = 0;i < n;i ++)
            cin >> ar[i];
        vctrl occ,emp;
        for(int i = 0;i < n;i ++){
            if(ar[i] == 1)
                occ.pb(i);
            else
                emp.pb(i);
        }
        int ln1,ln2;
        ln1 = occ.size();
        ln2 = emp.size();
        if(ln1 == 0){
            cout << "0\n";
            return 0;
        }
        ll dp[ln1][ln2];
        for(int i = 0;i < ln1;i ++){
            for(int j = 0;j < ln2;j ++)
                dp[i][j] = 1000000000;
        }
        dp[0][0] = abs(occ[0] - emp[0]);
        for(int i = 1;i < ln2;i ++)
            dp[0][i] = min(dp[0][i - 1],abs(occ[0] - emp[i]));
        for(int i = 1;i < ln1;i ++){
            for(int j = i;j < ln2;j ++)
                dp[i][j] = min(dp[i][j - 1],dp[i - 1][j - 1] + abs(occ[i]- emp[j]));
        }
        cout << dp[ln1 - 1][ln2 - 1] << "\n";

    }
    

    return 0;
}