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 java.io.*; 
import java.math.*;
import java.util.*;
import java.awt.Point;
 
public class Main {
	//static final long MOD = 998244353L;
	//static final long INF = 100000000000007L;
	static final long MOD = 1000000007L;
	//static final int INF = 1000000007;
	
	public static void main(String[] args) {
		FastScanner sc = new FastScanner();
		PrintWriter pw = new PrintWriter(System.out);
		int N = sc.ni();
		int[][] nums = new int[N-2][3];
		ArrayList<Integer>[] locs = new ArrayList[N];
		for (int i = 0; i < N; i++)
			locs[i] = new ArrayList<Integer>();
		for (int i = 0; i < N-2; i++) {
			for (int j = 0; j < 3; j++) {
				nums[i][j] = sc.ni()-1;
				locs[nums[i][j]].add(i);
			}
		}
		int[] ans = new int[N];
		Arrays.fill(ans,-1);
		loop:
		for (int i = 0; i < N; i++) {
			if (locs[i].size() == 1) {
				ans[0] = i;
				int row = locs[i].get(0);
				for (int a = 0; a < 3; a++) {
					if (locs[nums[row][a]].size() == 2) {
						ans[1] = nums[row][a];
						break;
					}
				}
				for (int j = i+1; j < N; j++) {
					if (locs[j].size() == 1) {
						ans[N-1] = j;
						row = locs[j].get(0);
						for (int a = 0; a < 3; a++) {
							if (locs[nums[row][a]].size() == 2) {
								ans[N-2] = nums[row][a];
								break;
							}
						}
						break loop;
					}
				}
			}
		}
		
		
		for (int i = 2; i < N-2; i++) {
			int row = -1;
			for (int loc: locs[ans[i-2]]) {
				if (locs[ans[i-1]].contains(loc) && (i == 2 || !locs[ans[i-3]].contains(loc))) {
					row = loc;
					break;
				}
			}
			for (int j = 0; j < 3; j++) {
				if (nums[row][j] != ans[i-2] && nums[row][j] != ans[i-1]) {
					ans[i] = nums[row][j];
					break;
				}
			}
		}
		
		for (int num: ans)
			pw.print((num+1) + " ");
		
		pw.close();
 	}
	
	static class DisjointSetUnion {
		public int[] parent;
		public int[] weight;
		public int count;

		public DisjointSetUnion(int nodes) {
			count = nodes;
			parent = new int[nodes];
			weight = new int[nodes];
			for (int i = 0; i < nodes; i++) {
				parent[i] = i;
				weight[i] = 1;
			}
		}
		//"find"
		public int root(int p) {
			while (p != parent[p]) {
				p = parent[p];
			}
			return p;
		}

		//"union"
		public void connect(int p, int q) {
			int rootP = root(p);
			int rootQ = root(q);
			if (rootP == rootQ) return;
			if (weight[rootP] < weight[rootQ]) {
				parent[rootP] = rootQ;
				weight[rootQ] += weight[rootP];
			} else {
				parent[rootQ] = rootP;
				weight[rootP] += weight[rootQ];
			}
			count--;
		}

		public boolean connected(int p, int q) {
			return root(p) == root(q);
		}
	}
    
    public static long power(long x, long y, long p) { 
        // Initialize result
        long res = 1;
         
        // Update x if it is more   
        // than or equal to p 
        x = x % p;  
      
        while (y > 0) 
        { 
            // If y is odd, multiply x 
            // with result 
            if((y & 1)==1) 
                res = (res * x) % p; 
      
            // y must be even now 
            // y = y / 2 
            y = y >> 1;  
            x = (x * x) % p;  
        } 
        return res; 
    }
    
    public static long dist(int[] point, int[] point2) {
    	return (long)(Math.pow((point2[1]-point[1]),2)+Math.pow((point2[0]-point[0]),2));
    }
    
    public static long gcd(long a, long b) {
    	if (b == 0)
    		return a;
    	else
    		return gcd(b,a%b);
    }
    
    public static int[][] sort(int[][] array) {
    	//Sort an array (immune to quicksort TLE)
		Random rgen = new Random();
		for (int i = 0; i < array.length; i++) {
		    int randomPosition = rgen.nextInt(array.length);
		    int[] temp = array[i];
		    array[i] = array[randomPosition];
		    array[randomPosition] = temp;
		}
		Arrays.sort(array, new Comparator<int[]>() {
			  @Override
        	  public int compare(int[] arr1, int[] arr2) {
				  return arr2[2]-arr1[2]; //descending order
	          }
		});
		return array;
	}
    
    public static long[][] sort(long[][] array) {
    	//Sort an array (immune to quicksort TLE)
		Random rgen = new Random();
		for (int i = 0; i < array.length; i++) {
		    int randomPosition = rgen.nextInt(array.length);
		    long[] temp = array[i];
		    array[i] = array[randomPosition];
		    array[randomPosition] = temp;
		}
		Arrays.sort(array, new Comparator<long[]>() {
			  @Override
			  //Descending order
        	  public int compare(long[] arr1, long[] arr2) {
				  if (arr2[0] < arr1[0])
					  return -1;
				  if (arr2[0] > arr1[0])
					  return 1;
				  return 0;
	          }
		});
		return array;
	}
    
    static class FastScanner { 
        BufferedReader br; 
        StringTokenizer st; 
  
        public FastScanner() { 
            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 ni() { 
            return Integer.parseInt(next()); 
        } 
  
        long nl() { 
            return Long.parseLong(next()); 
        } 
  
        double nd() { 
            return Double.parseDouble(next()); 
        } 
  
        String nextLine() { 
            String str = ""; 
            try { 
                str = br.readLine(); 
            } catch (IOException e) {
                e.printStackTrace(); 
            } 
            return str; 
        }
    }
}