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.
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.*;

public class B {

    static BufferedReader br;
    static long mod = 1000000000 + 7;
    static HashSet<Integer> p = new HashSet<>();

    public static void main(String[] args) throws Exception {
        br = new BufferedReader(new InputStreamReader(System.in));
        int tc = 1;
       // tc = cinI();
   while (tc-->0){
       int n=cinI();
       int[] ar=readArray(n,0,0);

       int ans=0;
       for(int i=0;i<n;i++){
           if(ar[i]!=1)continue;

             boolean lef=false;
           for(int j=i-1;j>=0;j--){
               if(ar[j]==0){
                   ar[j]=-1;
                   ans+=(i-j);
                   lef=true;
                  // System.out.println(i+" "+j);
                   break;
               }
           }
          // System.out.println(i+" "+ans);
           if(lef)continue;
           for(int j=i+1;j<n;j++){
               if(ar[j]==0){
                   ar[j]=-1;
                   ans+=(j-i);
           break;
               }
           }
           //System.out.println(i+" "+ans );
       }
       System.out.println(ans);
   }
    }

    public static void printDp(long[][] dp) {

        int n = dp.length;
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < dp[0].length; j++) {
                System.out.print(dp[i][j] + " ");
            }
            System.out.println();
        }
    }

    private static int dfs(TreeMap<Integer, HashSet<Integer>> graph, boolean[] vis, Integer key, Stack<Integer> st) {
        if (st.size() == 0) {
            return 0;
        }
        int par = st.pop();

        HashSet<Integer> child = graph.get(par);
        vis[par] = true;
        int nodes = 1;
        for (int c : child) {
            if (vis[c] == false) {
                st.add(c);
                nodes += dfs(graph, vis, c, st);
            }
        }
        return nodes;
    }


    public static boolean isSorted(int[] arr) {

        for (int i = 0; i < arr.length - 1; i++) {
            if (arr[i] > arr[i + 1]) {
                return false;
            }
        }
        return true;
    }


    private static long gcd(long a, long b) {
        if (a == 0)
            return b;
        if (b == 0)
            return a;

        // base case
        if (a == b)
            return a;

        // a is greater
        if (a > b)
            return gcd(a % b, b);
        return gcd(a, b % a);
    }


    public static long min(long a, long b) {
        return Math.min(a, b);
    }

    public static int min(int a, int b) {
        return Math.min(a, b);
    }


    public static void sieve() {
        int[] pf = new int[100000000 + 1];
        //0 prime //1 not prime
        pf[0] = 1;
        pf[1] = 1;
        for (int j = 2; j <= 10000; j++) {

            if (pf[j] == 0) {
                p.add(j);
                for (int k = j * j; k < pf.length; k += j) {
                    pf[k] = 1;
                }
            }
        }

    }


    public static int[] readArray(int n, int x, int z) throws Exception {
        int[] arr = new int[n];
        String[] ar = cinA();
        for (int i = x; i < n + x; i++) {
            arr[i] = getI(ar[i - x]);
        }
        return arr;
    }

    public static long[] readArray(int n, int x) throws Exception {
        long[] arr = new long[n];
        String[] ar = cinA();
        for (int i = x; i < n + x; i++) {
            arr[i] = getL(ar[i - x]);
        }
        return arr;
    }

    public static void arrinit(String[] a, long[] b) throws Exception {
        for (int i = 0; i < a.length; i++) {
            b[i] = Long.parseLong(a[i]);
        }
    }

    public static HashSet<Integer>[] Graph(int n, int edge, int directed) throws Exception {
        HashSet<Integer>[] tree = new HashSet[n];

        for (int j = 0; j < edge; j++) {

            String[] uv = cinA();
            int u = getI(uv[0]);
            int v = getI(uv[1]);
            if (directed == 0) {

                tree[v].add(u);
            }
            tree[u].add(v);
        }
        return tree;
    }

    public static void arrinit(String[] a, int[] b) throws Exception {
        for (int i = 0; i < a.length; i++) {
            b[i] = Integer.parseInt(a[i]);
        }
    }


    static double findRoots(int a, int b, int c) {
        // If a is 0, then equation is not
        //quadratic, but linear


        int d = b * b - 4 * a * c;
        double sqrt_val = Math.sqrt(Math.abs(d));


        // System.out.println("Roots are real and different \n");

        return Math.max((double) (-b + sqrt_val) / (2 * a),
                (double) (-b - sqrt_val) / (2 * a));


    }

    public static String cin() throws Exception {
        return br.readLine();
    }

    public static String[] cinA() throws Exception {
        return br.readLine().split(" ");
    }

    public static String[] cinA(int x) throws Exception {
        return br.readLine().split("");
    }

    public static String ToString(Long x) {
        return Long.toBinaryString(x);
    }

    public static void cout(String s) {
        System.out.println(s);
    }

    public static Integer cinI() throws Exception {
        return Integer.parseInt(br.readLine());
    }

    public static int getI(String s) throws Exception {
        return Integer.parseInt(s);
    }

    public static long getL(String s) throws Exception {
        return Long.parseLong(s);
    }

    public static long max(long a, long b) {
        return Math.max(a, b);
    }

    public static int max(int a, int b) {
        return Math.max(a, b);
    }

    public static void coutI(int x) {
        System.out.println(String.valueOf(x));
    }

    public static void coutI(long x) {
        System.out.println(String.valueOf(x));
    }

    public static Long cinL() throws Exception {
        return Long.parseLong(br.readLine());
    }

    public static void arrInit(String[] arr, int[] arr1) throws Exception {
        for (int i = 0; i < arr.length; i++) {
            arr1[i] = getI(arr[i]);
        }

    }
}