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>
//#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
#define zad tree<int, nuint_type, less<int>, rb_tree_tag, tree_order_statistics_node_update>
using namespace std;

#define in1(x) scanf("%lld",&x)
#define in2(x,y) scanf("%lld%lld",&x,&y)
#define in3(x,y,z) scanf("%lld%lld%lld",&x,&y,&z)
#define out(x) printf("%lld",x)
#define outspace(x) printf("%lld ",x)
#define outln(x) printf("%lld\n",x)

#define int long long
#define ld long double
#define pb push_back
#define p push
#define mkp make_pair
#define ff first
#define PLL pair<int,int>
#define PSS pair<string,string>
#define ss second
#define precision(x,d) cout<<fixed<<setprecision(d)<<x
#define minQueue priority_queue<int,vector<int>,greater<int> >
#define maxQueue priority_queue<int,vector<int>,less<int> >
#define all(x) x.begin(),x.end()
#define F(i,a,b) for(int i=a;i<b;i++)
#define RF(i,a,b) for(int i=a;i>b;i--)
#define mset(a,b) memset((a),(b),sizeof((a)))
#define IOS ios_base::sync_with_stdio(false),cin.tie(0),cout.tie(0);srand(time(NULL));
#define MOD 1000000007
#define RMOD 998244353
#define INF 10000000000000000
#define MINF LONG_LONG_MIN
#define MAXF LONG_LONG_MAX
const int N=5e3+5;
#define OuterLoop int t;scanf("%lld",&t);while(t--)
#define lcm(a,b) a*b/(__gcd(a,b))

int n;
vector<int> v,prefix;
int dp[N][N];

int go(int x,int y)
{
    if(x>=n) return 0;
    if(y>=n) 
    {
        if(prefix[x]==0) return 0;
        return INF;
    }

    int &ans = dp[x][y];
    if(ans!=-1) return ans;
    ans = INF;
    if(v[x])
    {
        if(!v[y])
        {
            ans = min(ans,abs(x-y)+go(x+1,y+1));
        }
        ans=min(ans,go(x,y+1));
    }
    else
    {
        ans = min(ans,go(x+1,y));
    }

    return ans;
}

void solve()
{
    cin>>n;
    v = vector<int>(n);
    prefix = vector<int>(n);
    for(int i=0;i<n;i++)
    {
        cin>>v[i];
    }

    for(int i=n-2;i>=0;i--)
    {
        prefix[i]=prefix[i+1]+v[i];
    }
    memset(dp,-1,sizeof(dp));
    int ans=go(0,0);
    cout<<ans<<"\n";
} 

int32_t main()
{
    IOS
    int t=1;
    //cin>>t;
    while(t--)
    {
        solve();
    }
}