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.File;
import java.io.FileNotFoundException;
import java.util.ArrayDeque;
import java.util.ArrayList;
import java.util.Scanner;
import java.util.StringTokenizer;

public class E {

	public static void main(String[] args) throws Exception {
//		StringTokenizer stok = new StringTokenizer(new Scanner(new File("F:/books/input.txt")).useDelimiter("\\A").next());
		StringTokenizer stok = new StringTokenizer(new Scanner(System.in).useDelimiter("\\A").next());
		StringBuilder sb = new StringBuilder();
		int n = Integer.parseInt(stok.nextToken());
		ArrayList<Integer>[] G = new ArrayList[n];
		for (int i = 0; i < n; i++)
			G[i] = new ArrayList<Integer>();
		for (int i = 0; i < n-1; i++) {
			int u = Integer.parseInt(stok.nextToken())-1;
			int v = Integer.parseInt(stok.nextToken())-1;
			G[u].add(v);
			G[v].add(u);
		}
		int[] a = dfsTree(G,0);
		int[][] b = new int[n+1][2];
		for(int i=0;i<a.length;i++) {
			if(b[a[i]][0]==0) b[a[i]][0] = i+1; else b[a[i]][1] = i+1;
		}
		for(int i=1;i<=n;i++) sb.append(b[i][0]+" "+b[i][1]+"\n");
		System.out.println(sb);
	}
	
	private static int[] dfsTree(ArrayList<Integer>[] G, int s) {
		int n = G.length;
		boolean[] visited= new boolean[n];
		int ptr=0;
		int[] ret = new int[2*n];
		int[] ss = new int[n];
		int[] p = new int[n];
		p[s] = s;
		ArrayDeque<Integer> S = new ArrayDeque<Integer>();
		S.add(s);
		while(!S.isEmpty()) {
			int c = S.pop();
			if(visited[c]!=true) {
				visited[c] = true;
				while(ss[c]<G[c].size() && visited[G[c].get(ss[c])]) ss[c]++;
				if(ss[c]==G[c].size()) {
					ret[ptr++]=c+1;
				}
				else {
					S.push(c);
					p[G[c].get(ss[c])]=c;
					S.push(G[c].get(ss[c]));
				}
			}
			else {
				ss[c]++;
				while(ss[c]<G[c].size() && visited[G[c].get(ss[c])]) ss[c]++;
				if(ss[c]==G[c].size()) {
					ret[ptr++] = c+1;
					for(int i=ss[c]-1;i>=0;i--) {
						if(G[c].get(i)!=p[c]) ret[ptr++] = G[c].get(i)+1;
					}
				}
				else {
					S.push(c);
					p[G[c].get(ss[c])]=c;
					S.push(G[c].get(ss[c]));
				}
			}
		}
		ret[ptr++]=s+1;
		return ret;
	}

}