Reconstruction of a Tree Aizu

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 n, 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 1 to n. Note that the root does not always correspond to 1.

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≤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

Code

/*
                                ^....0
                               ^ .1 ^1^
                               ..     01
                              1.^     1.0
                             ^ 1  ^    ^0.1
                             1 ^        ^..^
                             0.           ^ 0^
                             .0            1 .^
                             .1             ^0 .........001^
                             .1               1. .111100....01^
                             00                 11^        ^1. .1^
                             1.^                              ^0  0^
                               .^                                 ^0..1
                               .1                                   1..^
                             1 .0                                     ^  ^
                              00.                                     ^^0.^
                              ^ 0                                     ^^110.^
                          0   0 ^                                     ^^^10.01
                   ^^     10  1 1                                      ^^^1110.1
                   01     10  1.1                                      ^^^1111110
                   010    01  ^^                                        ^^^1111^1.^           ^^^
                   10  10^ 0^ 1                                            ^^111^^^0.1^       1....^
                    11     0                                               ^^11^^^ 0..  ....1^   ^ ^
                    1.     0^                                               ^11^^^ ^ 1 111^     ^ 0.
                   10   00 11                                               ^^^^^   1 0           1.
                   0^  ^0  ^0                                                ^^^^    0            0.
                   0^  1.0  .^                                               ^^^^    1 1          .0
                   ^.^  ^^  0^                             ^1                ^^^^     0.         ^.1
                   1 ^      11                             1.                ^^^     ^ ^        ..^
                  ^..^      ^1                             ^.^               ^^^       .0       ^.0
                  0..^      ^0                              01               ^^^       ..      0..^
                 1 ..        .1                             ^.^              ^^^       1 ^  ^0001
                ^  1.        00                              0.             ^^^        ^.0 ^.1
                . 0^.        ^.^                             ^.^            ^^^         ..0.0
               1 .^^.         .^                  1001        ^^            ^^^         . 1^
               . ^ ^.         11                0.    1         ^           ^^          0.
                0  ^.          0              ^0       1                   ^^^          0.
              0.^  1.          0^             0       .1                   ^^^          ..
              .1   1.          00            .        .1                  ^^^           ..
             1      1.         ^.           0         .^                  ^^            ..
             0.     1.          .^          .         0                                  .
             .1     1.          01          .        .                                 ^ 0
            ^.^     00          ^0          1.       ^                                 1 1
            .0      00           .            ^^^^^^                                   .
            .^      00           01                                                    ..
           1.       00           10                                                   1 ^
          ^.1       00           ^.                                            ^^^    .1
          ..        00            .1                                        1..01    ..
         1.1         00           1.                                       ..^      10
        ^ 1^         00           ^.1                                      0 1      1
        .1           00            00                                       ^  1   ^
         .           00            ^.^                                        10^  ^^
       1.1           00             00                                              10^
       ..^           1.             ^.                                               1.
      0 1            ^.              00                 00                            .^
        ^            ^.              ^ 1                00   ^0000^     ^               01
     1 0             ^.               00.0^              ^00000   1.00.1              11
     . 1              0               1^^0.01                      ^^^                01
      .^              ^                1   1^^                                       ^.^
    1 1                                                                              0.
    ..                                                                              1 ^
     1                                                                               1
   ^ ^                                                                             .0
   1                                                                             ^ 1
   ..                                                          1.1            ^0.0
  ^ 0                                                           1..01^^100000..0^
  1 1                                                            ^ 1 ^^1111^ ^^
  0 ^                                                             ^ 1      1000^
  .1                                                               ^.^     .   00
  ..                                                                1.1    0.   0
  1.                                                                  .    1.   .^
  1.                                                                 1    1.   ^0
 ^ .                                                                 ^.1 00    01
 ^.0                                                                  001.     .^
 */
// Virtual_Judge —— Reconstruction of a Tree Aizu - ALDS1_7_d.cpp created by VB_KoKing on 2019-05-09:10.
/* Procedural objectives：

 Variables required by the program:

 Procedural thinking：

 Functions required by the program:
 
 Determination algorithm:
 
 Determining data structure:
 

*/
/* My dear Max said：
"I like you,
So the first bunch of sunshine I saw in the morning is you,
The first gentle breeze that passed through my ear is you,
The first star I see is also you.
The world I see is all your shadow."

FIGHTING FOR OUR FUTURE！！！
*/
#include <algorithm>
#include <iostream>
#include <vector>

using namespace std;

int pos = 0;
vector<int> pre, in, post;

void rec(int left, int right) {
    if (left >= right) return;
    int root = pre[pos++];
    int m = distance(in.begin(), find(in.begin(), in.end(), root));
    rec(left, m);
    rec(m + 1, right);
    post.push_back(root);
}

void solve() {
    rec(0, pre.size());
    for (int i = 0; i < post.size(); i++) {
        if (i) cout << ' ';
        cout << post[i];
    }
    cout << endl;
}

int main() {
    int n;
    cin >> n;
    for (int i = 0; i < n; i++) {
        int k;
        cin >> k;
        pre.push_back(k);
    }
    for (int i = 0; i < n; i++) {
        int k;
        cin >> k;
        in.push_back(k);
    }
    solve();
    return 0;
}