Problem: We had a really tough time generating tests for problem D. In order to prepare strong tests, we had to solve the following problem.

Given an undirected labeled tree consisting of n vertices, find a set of segments such that:

  1. both endpoints of each segment are integers from 1 to 2n, and each integer from 1 to 2n should appear as an endpoint of exactly one segment; 
  2. all segments are non-degenerate; 
  3. for each pair (i, j) such that i ≠ j, i ∈ [1, n] and j ∈ [1, n], the vertices i and j are connected with an edge if and only if the segments i and j intersect, but neither segment i is fully contained in segment j, nor segment j is fully contained in segment i. 



Can you solve this problem too?

Input

The first line contains one integer n (1 ≤ n ≤ 5 ⋅ 10^5) — the number of vertices in the tree.

Then n - 1 lines follow, each containing two integers x_i and y_i (1 ≤ x_i, y_i ≤ n, x_i ≠ y_i) denoting the endpoints of the i-th edge.

It is guaranteed that the given graph is a tree.

Output

Print n pairs of integers, the i-th pair should contain two integers l_i and r_i (1 ≤ l_i < r_i ≤ 2n) — the endpoints of the i-th segment. All 2n integers you print should be unique.

It is guaranteed that the answer always exists.

Examples

Input


6
1 2
1 3
3 4
3 5
2 6


Output


9 12
7 10
3 11
1 5
2 4
6 8


Input


1


Output


1 2

Solution (in JAVA):

Solution:
import java.io.*;
import java.math.BigInteger;
import java.util.*;


 
public class Main{
	public static class FastReader {
		BufferedReader br;
		StringTokenizer root;
		
 
		public FastReader() {
			br = new BufferedReader(new InputStreamReader(System.in));
		}
 
		String next() {
			while (root == null || !root.hasMoreTokens()) {
				try {
					root = new StringTokenizer(br.readLine());
				} catch (Exception r) {
					r.printStackTrace();
				}
			}
			return root.nextToken();
		}
 
		int nextInt() {
			return Integer.parseInt(next());
		}
 
		double nextDouble() {
			return Double.parseDouble(next());
		}
 
		long nextLong() {
			return Long.parseLong(next());
		}
 
		String nextLine() {
			String str = "";
			try {
				str = br.readLine();
			} catch (Exception r) {
				r.printStackTrace();
			}
			return str;
		}
	}
	
	public static PrintWriter out = new PrintWriter (new BufferedOutputStream(System.out));
	static long mod = (long) (1e9+7);
	static long cf = 998244353;
    static final long MAX = (long) 1e8;
    public static List<Integer>[] edges;
    public static int[][] parent;
    public static long[] fac;
    public static int N = (int) 1e5;
    public static int x = 0;
	public static void main(String[] args) {
		FastReader sc = new FastReader();
        
		int n = sc.nextInt();
		edges = new ArrayList[n+1];
		for(int i=0;i<=n;++i) edges[i] = new ArrayList<>();
		
		for(int i=0;i<n-1;++i) {
			int u = sc.nextInt();
			int v = sc.nextInt();
			edges[u].add(v);
			edges[v].add(u);
		}
		int[] l = new int[n+1];
		int[] r = new int[n+1];
		l[1] = ++x;
		dfs(1,0,l,r);
		for(int i=1;i<=n;++i) out.println(l[i] +" "+ r[i]);
	
		out.close();
}
	private static void dfs(int v, int p, int[] l, int[] r) {
		for(int child : edges[v]) {
			if(child != p) {
				l[child] = ++x; // add each child
			}
		}
		// close parent
		r[v] = ++x;
		for(int i = edges[v].size()-1;i>=0;--i) {
			int child = edges[v].get(i);
			if(child != p)
				dfs(child,v,l,r);
		}
		
		
	}
	
}