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.util.*;

public class E
{
	static ArrayList<Integer>[] adj;
	static int[] stk;
	static int ptr;

	static void dfs(int u, int par)
	{
		for (int i = adj[u].size() - 1; i >= 0; i--)
		{
			int v = adj[u].get(i);
			if (v != par)
				stk[ptr++] = v;
		}

		stk[ptr++] = -u;

		for (int v : adj[u])
			if (v != par)
				dfs(v, u);
	}

	static void solve(FastIO io)
	{
		int n = io.nextInt();
		adj = new ArrayList[n + 1];

		for (int i = 0; i <= n; i++)
			adj[i] = new ArrayList<>();
		
		for (int i = 0; i < n - 1; i++)
		{
			int u = io.nextInt();
			int v = io.nextInt();

			adj[u].add(v);
			adj[v].add(u);
		}
		
		stk = new int[2*n];
		stk[ptr++] = 1;
		dfs(1, -1);

		// io.println(list);

		int[] left = new int[n+1];
		int[] right = new int[n+1];
		int idx = 1;

		for (int i = 2*n - 1; i >= 0; i--)
		{
			int k = stk[i];

			if (k < 0)
				left[-1 * k]= idx++;
			else
				right[k] = idx++;
		}

		for (int i = 1; i <= n; i++)
			io.println(left[i] + " " + right[i]);
	}

	public static void main(String[] args) {
		FastIO io = new FastIO();

		solve(io);

		io.close();
	}
}

class FastIO extends PrintWriter
{
	BufferedReader r = new BufferedReader(new InputStreamReader(System.in));
	StringTokenizer st = new StringTokenizer("");

	FastIO()
	{
		super(System.out);
	}

	public String next()
	{
		while (!st.hasMoreTokens())
		{
			try {
				st = new StringTokenizer(r.readLine());
			} catch (Exception e) {
				//TODO: handle exception
			}
		}
		return st.nextToken();
	}

	public int nextInt()
	{
		return Integer.parseInt(next());
	}

	public long nextLong()
	{
		return Long.parseLong(next());
	}

	public double nextDouble()
	{
		return Double.parseDouble(next());
	}
}