Vasya has got n books, numbered from 1 to n, arranged in a stack. The topmost book has number a_1, the next one — a_2, and so on. The book at the bottom of the stack has number a_n. All numbers are distinct.

Vasya wants to move all the books to his backpack in n steps. During i-th step he wants to move the book number b_i into his backpack. If the book with number b_i is in the stack, he takes this book and all the books above the book b_i, and puts them into the backpack; otherwise he does nothing and begins the next step. For example, if books are arranged in the order [1, 2, 3] (book 1 is the topmost), and Vasya moves the books in the order [2, 1, 3], then during the first step he will move two books (1 and 2), during the second step he will do nothing (since book 1 is already in the backpack), and during the third step — one book (the book number 3). Note that b_1, b_2, ..., b_n are distinct.

Help Vasya! Tell him the number of books he will put into his backpack during each step.

Input

The first line contains one integer n~(1 ≤ n ≤ 2 ⋅ 10^5) — the number of books in the stack.

The second line contains n integers a_1, a_2, ..., a_n~(1 ≤ a_i ≤ n) denoting the stack of books.

The third line contains n integers b_1, b_2, ..., b_n~(1 ≤ b_i ≤ n) denoting the steps Vasya is going to perform.

All numbers a_1 ... a_n are distinct, the same goes for b_1 ... b_n.

Output

Print n integers. The i-th of them should be equal to the number of books Vasya moves to his backpack during the i-th step.

Examples

Input

3
1 2 3
2 1 3


Output

2 0 1 


Input

5
3 1 4 2 5
4 5 1 3 2


Output

3 2 0 0 0 


Input

6
6 5 4 3 2 1
6 5 3 4 2 1


Output

1 1 2 0 1 1 

Note

The first example is described in the statement.

In the second example, during the first step Vasya will move the books [3, 1, 4]. After that only books 2 and 5 remain in the stack (2 is above 5). During the second step Vasya will take the books 2 and 5. After that the stack becomes empty, so during next steps Vasya won't move any books.
import java.io.*;
import java.util.Arrays;
import java.util.StringTokenizer;

public class Pro2_10_24 {
    public static void main(String[] args) {

        FastReader in = new FastReader();
        OutputStream outputStream = System.out;
        PrintWriter out = new PrintWriter(outputStream);
        int i,j,a,b,c,n,k,l,m;
        String s,s1,s2;
        boolean is1 = false ,is = false,was = false;
        long sum = 0;
        n = in.nextInt();
        int[] brr = new int[n+1];
        int[] arr = new int[n+1];
        int[] crr = new int[n+1];
        //mergeSort(arr, 0, 7);
        //Arrays.binarySearch(arr,7);

        for (i=1; i<=n; i++){
            arr[i] = in.nextInt();
            brr[arr[i]] = i;
        }

        for (i=1; i<=n; i++){
            crr[i] = in.nextInt();
        }

        k = 0;
        for (i=1; i<= n; i++){
            if ( arr[brr[crr[i]]] == 0) {
                out.print(0+ " ");
                out.flush();
                continue;
            }

            out.print( (brr[crr[i]] - sum) + " ");

            sum = Math.max(sum,brr[crr[i]]);

            for (j=brr[crr[i]]; j>0; j--){
                if (arr[j] == 0) break;
                    arr[j] = 0;
                    k++;
            }
        }

        //out.println(k);

        out.flush();


    }


    static int GCD(int a, int b) {
        return a == 0 ? b : GCD(b % a, a);
    }


    static void  mergeSort(int arr[], int l, int r)
    {
        if (l < r)
        {
            // Same as (l+r)/2, but avoids overflow for
            // large l and h
            int m = l+(r-l)/2;

            // Sort first and second halves
            mergeSort(arr, l, m);
            mergeSort(arr, m+1, r);

            merge(arr, l, m, r);
        }
    }


    static void merge(int arr[], int l, int m, int r)
    {
        int i, j, k;
        int n1 = m - l + 1;
        int n2 =  r - m;

        /* create temp arrays */
        int L[] = new int[n1],  R[] = new int[n2];

        /* Copy data to temp arrays L[] and R[] */
        for (i = 0; i < n1; i++)
            L[i] = arr[l + i];
        for (j = 0; j < n2; j++)
            R[j] = arr[m + 1+ j];

        /* Merge the temp arrays back into arr[l..r]*/
        i = 0; // Initial index of first subarray
        j = 0; // Initial index of second subarray
        k = l; // Initial index of merged subarray
        while (i < n1 && j < n2)
        {
            if (L[i] <= R[j])
            {
                arr[k] = L[i];
                i++;
            }
            else
            {
                arr[k] = R[j];
                j++;
            }
            k++;
        }

    /* Copy the remaining elements of L[], if there
       are any */
        while (i < n1)
        {
            arr[k] = L[i];
            i++;
            k++;
        }

    /* Copy the remaining elements of R[], if there
       are any */
        while (j < n2)
        {
            arr[k] = R[j];
            j++;
            k++;
        }
    }






    static class FastReader {
        BufferedReader br;
        StringTokenizer st;

        public FastReader() {
            br = new BufferedReader(new
                    InputStreamReader(System.in));
        }

        String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        int nextInt() {
            return Integer.parseInt(next());
        }

        long nextLong() {
            return Long.parseLong(next());
        }

        double nextDouble() {
            return Double.parseDouble(next());
        }

        String nextLine() {
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
    }

}
