Write a program which reads two sequences of nodes obtained by the preorder tree walk and the inorder tree walk on a binary tree respectively, and prints a sequence of the nodes obtained by the postorder tree walk on the binary tree.

Input

In the first line, an integer nn, which is the number of nodes in the binary tree, is given.
In the second line, the sequence of node IDs obtained by the preorder tree walk is given separated by space characters.
In the second line, the sequence of node IDs obtained by the inorder tree walk is given separated by space characters.

Every node has a unique ID from 11 to nn. Note that the root does not always correspond to 11.

Output

Print the sequence of node IDs obtained by the postorder tree walk in a line. Put a single space character between adjacent IDs.

Constraints

• 1≤n≤401≤n≤40

Sample Input 1

5
1 2 3 4 5
3 2 4 1 5

Sample Output 1

3 4 2 5 1

Sample Input 2

4
1 2 3 4
1 2 3 4

Sample Output 2

4 3 2 1

 給出先序遍歷和中序遍歷，求後序遍歷。

可以通過遍歷先序，確定節點的左右子樹，然後求解。

程式碼如下：
 

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <vector>
using namespace std;
int n;
int loc=0;
vector <int> pre,in,post;
int finds (int x)
{
    for (int i=0;i<n;i++)
        if(in[i]==x)
           return i;
}
void rec (int l,int r)
{
    if(l>=r)
        return;
    int root=pre[loc++];
    int m=finds(root);
    rec(l,m);
    rec(m+1,r);
    post.push_back(root);
}
void output ()
{
    rec (0,n);
    for (int i=0;i<n;i++)
        printf("%d%c",post[i],i==n-1?'\n':' ');
}
int main()
{
    scanf("%d",&n);
    for (int i=0;i<n;i++)
    {
        int x;
        scanf("%d",&x);
        pre.push_back(x);
    }
    for (int i=0;i<n;i++)
    {
        int x;
        scanf("%d",&x);
        in.push_back(x);
    }
    output();
    return 0;
}