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.util.*;
import java.io.*;
public class Main{
    static PrintWriter out;
    static InputReader in;
    public static void main(String args[]){
        out = new PrintWriter(System.out);
        in = new InputReader();
        new Main();
        out.flush(); out.close();
    }   
    Main(){
        solve();
    }
    class pair{
        int F, S;
        pair(int a, int b){
            F = a; S = b;
        }
    }
    final int max = 500010;
    int val[] = new int[2 * max];
    ArrayList<Integer> al[] = new ArrayList[max];
    int p = 0;
    void dfs(int u, int pa){
        for(int v : al[u]){
            if(v == pa)continue;
            dfs(v, u);
        }
        val[p++] = u;
        for(int i = al[u].size() - 1; i >= 0; i--){
            int v = al[u].get(i);
            if(v == pa)continue;
            val[p++] = v;
        }
    }
    void solve(){
        int n = in.nextInt();
        for(int i = 0; i <= n; i++)al[i] = new ArrayList<>();
        for(int i = 0; i < n - 1; i++){
            int u = in.nextInt(), v = in.nextInt();
            al[u].add(v); al[v].add(u);
        }
        dfs(1, 0);
        val[p++] = 1;
        int dp[][] = new int[n + 1][2];
        for(int i = 0; i < p; i++){
            int node = val[i];
            // System.out.print(node + " ");
            if(dp[node][0] != 0)dp[node][1] = i + 1;
            else dp[node][0] = i + 1;
        }
        for(int i = 1; i <= n; i++){
            out.println(dp[i][0] + " " + dp[i][1]);
        }
    }   
    public static class InputReader{
        BufferedReader br;
        StringTokenizer st;
        InputReader(){
            br = new BufferedReader(new InputStreamReader(System.in));
        }
        public int nextInt(){
            return Integer.parseInt(next());
        }
        public long nextLong(){
            return Long.parseLong(next());
        }
        public double nextDouble(){
            return Double.parseDouble(next());
        }
        public String next(){
            while(st == null || !st.hasMoreTokens()){
                try{
                    st = new StringTokenizer(br.readLine());
                }catch(IOException e){}
            }
            return st.nextToken();
        }
    }
}