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>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <cmath>
#include <vector>
#include <set>
#include <map>
#include <unordered_set>
#include <unordered_map>
#include <queue>
#include <ctime>
#include <cassert>
#include <complex>
#include <string>
#include <cstring>
#include <chrono>
#include <random>
#include <queue>
#include <bitset>
using namespace std;

typedef long long ll;
typedef long long unsigned int llui;
typedef pair<int,int> ii;
typedef pair<ll,ll> pll;
typedef vector<ii> vii;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef vector< vector<int> > vvi;
typedef vector< vector<ll> > vvl;
typedef set<int> si;
typedef map<string, int> msi;
typedef map<int, int> mii;

using str = string; // yay python!

#define INF 1000000000
#define MOD 1000000007
#define EPS 1e-9
#define ALL(x) (x).begin(), (x).end()
#define RALL(x) (x).rbegin(), (x).rend()
#define SORT(x) sort((x).begin(),(x).end())
#define UNIQUE(x) SORT(x),(x).erase(unique((x).begin(),(x).end()),(x).end())
#define F first
#define S second
#define MP make_pair
#define PB push_back
#define EB emplace_back
#define LB lower_bound
#define UB upper_bound
#define SZ(a) int((a).size())
#define CLR(a) memset(a,0,sizeof(a))
#define SET(a,b) memset(a,b,sizeof(a))
#define POSL(x,v) (lower_bound(x.begin(),x.end(),v)-x.begin())
#define POSU(x,v) (upper_bound(x.begin(),x.end(),v)-x.begin())

#define FOR(i,a,b) for (int i = int(a); i < int(b); i++)
#define FORD(i,a,b) for(int i=int(b)-1; i>=int(a); i--)
#define REP(i, n) for (int i = 0; i < int(n); i++)
#define REP1(i,n) for (int i = 1; i <= int(n); i++)
#define REV(i, n) for (int i = int(n)-1; i >=0 ; i--)

#define sd(n) scanf("%d",&n)
#define dsd(n) int n; scanf("%d",&n)
#define sd2(a,b) scanf("%d%d",&a,&b)
#define dsd2(a,b) int a,b; scanf("%d%d",&a,&b)
#define sd3(a,b,c) scanf("%d%d%d",&a,&b,&c)
#define dsd3(a,b,c) int a,b,c; scanf("%d%d%d",&a,&b,&c)
#define pd(n) printf("%d\n",n)
#define pds(n) printf("%d ",n)
#define sl(n) scanf("%lld",&n)
#define dsl(n) ll n; scanf("%lld",&n)
#define sl2(a,b) scanf("%lld%lld",&a,&b)
#define dsl2(a,b) ll a,b; scanf("%lld%lld",&a,&b)
#define sl3(a,b,c) scanf("%lld%lld%lld",&a,&b,&c)
#define dsl3(a,b,c) ll a,b,c; scanf("%lld%lld%lld",&a,&b,&c)
#define pl(n) printf("%lld\n",n)  
#define pls(n) printf("%lld ",n)  

#define sim template < class c
#define ris return * this
#define dor > debug & operator <<
#define eni(x) sim > typename \
  enable_if<sizeof dud<c>(0) x 1, debug&>::type operator<<(c i) {
sim > struct rge { c b, e; };
sim > rge<c> range(c i, c j) { return rge<c>{i, j}; }
sim > auto dud(c* x) -> decltype(cerr << *x, 0);
sim > char dud(...);
struct debug {
#ifdef LOCAL
~debug() { cerr << endl; }
eni(!=) cerr << boolalpha << i; ris; }
eni(==) ris << range(begin(i), end(i)); }
sim, class b dor(pair < b, c > d) {
  ris << "(" << d.first << ", " << d.second << ")";
}
sim dor(rge<c> d) {
  *this << "[";
  for (auto it = d.b; it != d.e; ++it)
    *this << ", " + 2 * (it == d.b) << *it;
  ris << "]";
}
#else
sim dor(const c&) { ris; }
#endif
};
#define imie(...) " [" << #__VA_ARGS__ ": " << (__VA_ARGS__) << "] "

inline int two(int n) { return 1 << n; }
inline int test(int n, int b) { return (n>>b)&1; }
inline void set_bit(int & n, int b) { n |= two(b); }
inline void unset_bit(int & n, int b) { n &= ~two(b); }
inline int last_bit(int n) { return n & (-n); }
inline int ones(int n) { int res = 0; while(n && ++res) n-=n&(-n); return res; }

#define LSOne(S) (S&(-S))

const double pi = acos(-1);
//const int oo = 0x3f3f3f3f;

int cmp(double x, double y = 0, double tol = EPS) {
    return (x <= y + tol) ? (x + tol < y) ? -1 : 0 : 1;
}

int gcd(int a, int b) { return b ? gcd(b, a%b) : a; }

template<class T> bool uin(T &a,T b) {return a > b ? (a=b, true):false;}
template<class T> bool uax(T &a,T b) {return a < b ? (a=b, true):false;}

/////////////////////////////////////////////////////////////////////

const int nax = 5005;

int dp[nax][nax];

int main(){
    dsd(n);
    int N = n;
    vi V(n);
    REP(i,n) sd(V[i]);
    vi empty;
    vi pos;
    REP(i,n){
        if(V[i]==0) empty.PB(i);
        else pos.PB(i);
    }
    if(SZ(pos)==0) return !pd(0);
    REP(i, n) REP(j, n) dp[i][j] = INF;
    for(int i: empty){
        dp[0][i] = abs(i-pos[0]);
    }
    for(int i=1;i<N;i++) uin(dp[0][i], dp[0][i-1]);
    for(int i=1;i<SZ(pos);i++){
        for(int j:empty){
            if(j==0) continue;
            dp[i][j] = dp[i-1][j-1] + abs(j-pos[i]);
        }
        for(int j=1;j<N;j++){
            uin(dp[i][j], dp[i][j-1]);
        }
    }
    int ans = INF;
    REP(i, N) uin(ans, dp[SZ(pos)-1][i]);
    pd(ans);
    return 0;
}
// Do something instead of nothing, stay organised, stay ahead!, WRITE STUFF DOWN
