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.BufferedReader;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.StringTokenizer;

public class E {
	static PrintWriter out = new PrintWriter(System.out);

	static int[] l, r, sz;
	static ArrayList<Integer> adj[];
	static int N;
	
	public static void main(String args[]){
		new Thread(null, ()->asdf(args), "", 1<<28).start();
	}
	public static void asdf(String[] args) {
		FS in = new FS();
		N = in.nextInt();
		l = new int[N];
		r = new int[N];
		adj = new ArrayList[N];
		for(int i = 0; i < N; i++) {
			adj[i] = new ArrayList<Integer>();
		}
		for(int i = 0; i < N-1; i++) {
			int a = in.nextInt()-1;
			int b = in.nextInt()-1;
			adj[a].add(b);
			adj[b].add(a);
		}
		sz = new int[N];
		getSz(0, -1);
		l = new int[N];
		r = new int[N];
		dfs(0, -1, 1, 2*N);
		
		for(int i = 0; i < N; i++) {
			out.println(l[i]+" "+r[i]);
		}
		out.close();
	}
	
	static void getSz(int node, int p) {
		sz[node] = 1;
		for(int ii : adj[node]) {
			if(ii != p) { getSz(ii, node);
			sz[node] += sz[ii];
			}
		}
	}
	static void dfs(int node, int p, int ll, int rr) {
		if(p == -1) {
			l[node] = ll++;
			for(int ii : adj[node]) if(ii != p) r[ii] = rr--;
			r[node] = rr--;
			for(int ii : adj[node]) {
				if(ii != p) {
					int give = 2*sz[ii]-1;
					dfs(ii, node, ll, ll+give-1);
					ll += give;
				}
			}
		}
		else {
			for(int ii : adj[node]) if(ii != p) r[ii] = rr--;
			l[node] = rr--;
			for(int ii : adj[node]) {
				if(ii != p) {
					int give = 2*sz[ii]-1;
					dfs(ii, node, ll, ll+give-1);
					ll += give;
				}
			}
		}
		
	}
	
	static class FS{
		BufferedReader br;
		StringTokenizer st;
		public FS() {
			br = new BufferedReader(new InputStreamReader(System.in));
		}
		String next() {
			while(st == null || !st.hasMoreElements()) {
				try {st = new StringTokenizer(br.readLine());}
				catch(Exception e) { throw null;}
			}
			return st.nextToken();
		}
		int nextInt() { return Integer.parseInt(next());}
		double nextDouble() { return Double.parseDouble(next());}
		long nextLong() { return Long.parseLong(next());}
		int[] NIA(int n) {
			int r[] = new int[n];
			for(int i = 0; i < n; i++) r[i] = nextInt();
			return r;
		}
		long[] NLA(int n) {
			long r[] = new long[n];
			for(int i = 0; i < n; i++) r[i] = nextLong();
			return r;
		}
		char[][] grid(int r, int c){
			char res[][] = new char[r][c];
			for(int i = 0; i < r; i++) {
				char l[] = next().toCharArray();
				for(int j = 0; j < c; j++) {
					res[i][j] = l[j];
				}
			}
			return res;
		}
	}
	
}