Bob is an avid fan of the video game "League of Leesins", and today he celebrates as the League of Leesins World Championship comes to an end! 

The tournament consisted of n (n ≥ 5) teams around the world. Before the tournament starts, Bob has made a prediction of the rankings of each team, from 1-st to n-th. After the final, he compared the prediction with the actual result and found out that the i-th team according to his prediction ended up at the p_i-th position (1 ≤ p_i ≤ n, all p_i are unique). In other words, p is a permutation of 1, 2, ..., n.

As Bob's favorite League player is the famous "3ga", he decided to write down every 3 consecutive elements of the permutation p. Formally, Bob created an array q of n-2 triples, where q_i = (p_i, p_{i+1}, p_{i+2}) for each 1 ≤ i ≤ n-2. Bob was very proud of his array, so he showed it to his friend Alice.

After learning of Bob's array, Alice declared that she could retrieve the permutation p even if Bob rearranges the elements of q and the elements within each triple. Of course, Bob did not believe in such magic, so he did just the same as above to see Alice's respond.

For example, if n = 5 and p = [1, 4, 2, 3, 5], then the original array q will be [(1, 4, 2), (4, 2, 3), (2, 3, 5)]. Bob can then rearrange the numbers within each triple and the positions of the triples to get [(4, 3, 2), (2, 3, 5), (4, 1, 2)]. Note that [(1, 4, 2), (4, 2, 2), (3, 3, 5)] is not a valid rearrangement of q, as Bob is not allowed to swap numbers belong to different triples.

As Alice's friend, you know for sure that Alice was just trying to show off, so you decided to save her some face by giving her any permutation p that is consistent with the array q she was given. 

Input

The first line contains a single integer n (5 ≤ n ≤ 10^5) — the size of permutation p.

The i-th of the next n-2 lines contains 3 integers q_{i, 1}, q_{i, 2}, q_{i, 3} (1 ≤ q_{i, j} ≤ n) — the elements of the i-th triple of the rearranged (shuffled) array q_i, in random order. Remember, that the numbers within each triple can be rearranged and also the positions of the triples can be rearranged.

It is guaranteed that there is at least one permutation p that is consistent with the input. 

Output

Print n distinct integers p_1, p_2, …, p_n (1 ≤ p_i ≤ n) such that p is consistent with array q. 

If there are multiple answers, print any. 

Example

Input


5
4 3 2
2 3 5
4 1 2


Output


1 4 2 3 5 
import javax.print.DocFlavor;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.nio.charset.IllegalCharsetNameException;
import java.sql.SQLOutput;
import java.util.*;



public class Main {
    static LinkedList[] g;
    static int[] parent;

    public static void main(String[] args) throws IOException{
        Reader.init(System.in);
        int n = Reader.nextInt();
        HashMap<TreeSet<Integer>,Boolean> map = new HashMap<>();
        int[] cnt = new int[n+1];
        LinkedList<TreeSet<Integer>> arr[] = new LinkedList[n+1];
        for (int i = 0 ; i < n+1 ; i++) arr[i] = new LinkedList<>();
        for (int i = 0 ; i < n-2 ; i++) {
            TreeSet<Integer> set = new TreeSet<>();
            int num1 = Reader.nextInt();
            int num2 = Reader.nextInt();
            int num3 = Reader.nextInt();
            cnt[num1]++;
            cnt[num2]++;
            cnt[num3]++;
            set.add(num1);
            set.add(num2);
            set.add(num3);
            arr[num1].addFirst(set);
            arr[num2].addFirst(set);
            arr[num3].addFirst(set);
            map.put(set,true);
        }

        int first = -1;
        for (int i = 1 ; i<= n ; i++){
            if (cnt[i]==1){
                first = i;
                break;
            }
        }
        LinkedList<TreeSet<Integer>> lf = arr[first];
        TreeSet<Integer> tmp = lf.getFirst();
        Iterator<Integer> ii = tmp.iterator();
        int second = -1;
        while (ii.hasNext()){
            int nn = ii.next();
            if (cnt[nn]==2){
                second = nn;
                break;
            }
        }
        System.out.print(first + " " + second + " ");
        boolean[] vis = new boolean[n+1];
        vis[first] = true;
        vis[second] = true;
        for (int i = 0 ; i < n-2 ; i++){
            LinkedList<TreeSet<Integer>> temp = arr[second];
            Iterator my = temp.iterator();
            while (my.hasNext()){
                TreeSet<Integer> ok = (TreeSet<Integer>) my.next();
                if (ok.contains(first) && ok.contains(second)){
                    Iterator x = ok.iterator();
                    int f = 0;
                    while (x.hasNext()){
                        int number = (int) x.next();
                        if (number!= first && number!=second && vis[number]==false){
                            System.out.print(number + " ");
                            vis[number] = true;
                            first = second;
                            second = number;
                            f = 1;
                            break;
                        }
                    }
                    if (f==1){
                        break;
                    }
                }
            }

        }



    }

    static int find(int x) {
        return parent[x] == x ? x : (parent[x] = find(parent[x]));
    }
    static void merge(int x, int y) {
        x = find(x);
        y = find(y);
        if ( x > y ){
            parent[y] = x;
        }
        else if (y > x){
            parent[x] = y;
        }
    }

    static void solve()throws IOException{




    }

    public static void sortbyColumn(int arr[][], int col)
    {
        // Using built-in sort function Arrays.sort
        Arrays.sort(arr, new Comparator<int[]>() {

            @Override
            // Compare values according to columns
            public int compare(final int[] entry1,
                               final int[] entry2) {

                // To sort in descending order revert
                // the '>' Operator
                if (entry1[col] > entry2[col])
                    return 1;
                else
                    return -1;
            }
        });  // End of function call sort().
    }


    static class Node{
        long a;
        int i;
        Node(long a, int i){
            this.a = a;
            this.i = i;
        }
    }

}


class Reader {
    static BufferedReader reader;
    static StringTokenizer tokenizer;

    /** call this method to initialize reader for InputStream */
    static void init(InputStream input) {
        reader = new BufferedReader(
                new InputStreamReader(input) );
        tokenizer = new StringTokenizer("");
    }
    /** get next word */
    static String next() throws IOException {
        while ( ! tokenizer.hasMoreTokens() ) {
            //TODO add check for eof if necessary
            tokenizer = new StringTokenizer(
                    reader.readLine() );
        }
        return tokenizer.nextToken();
    }

    static int nextInt() throws IOException {
        return Integer.parseInt( next() );
    }
    static long nextLong() throws IOException {
        return Long.parseLong( next() );
    }

    static double nextDouble() throws IOException {
        return Double.parseDouble( next() );
    }
}
class MergeSort
{
    // Merges two subarrays of arr[].
    // First subarray is arr[l..m]
    // Second subarray is arr[m+1..r]
    void merge(int arr[], int l, int m, int r)
    {
        // Find sizes of two subarrays to be merged
        int n1 = m - l + 1;
        int n2 = r - m;

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

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


        /* Merge the temp arrays */

        // Initial indexes of first and second subarrays
        int i = 0, j = 0;

        // Initial index of merged subarry array
        int k = l;
        while (i < n1 && j < n2)
        {
            if (L[i] <= R[j])
            {
                arr[k] = L[i];
                i++;
            }
            else
            {
                arr[k] = R[j];
                j++;
            }
            k++;
        }

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

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

    // Main function that sorts arr[l..r] using
    // merge()
    void sort(int arr[], int l, int r)
    {
        if (l < r)
        {
            // Find the middle point
            int m = (l+r)/2;

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

            // Merge the sorted halves
            merge(arr, l, m, r);
        }
    }

    /* A utility function to print array of size n */
    static void printArray(int arr[])
    {
        int n = arr.length;
        for (int i=0; i<n; ++i)
            System.out.print(arr[i] + " ");
        System.out.println();
    }

    // Driver method

}

class Node implements Comparable<Node>{
    int a;
    int b;
    Node (int a , int b){
        this.a = a;
        this.b = b;
    }



    public int compareTo(Node o) {
        if ((this.a%2) == (o.a%2)){
            return (this.b - o.b);
        }
        else{
            return this.a - o.a;
        }
    }
}