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 CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int MX = 5e5 + 2;
vector<int> adj[MX];
pair<int, int> ans[MX];
int ccnt[MX];
int dfs(int i, int p, int l, int r) {
  ans[i] = {l, r};
  int bnd = r, ptr = r - 1;
  for (int a : adj[i]) {
    if (a == p) continue;
    bnd = dfs(a, i, ptr, bnd + (int)adj[a].size());
    ptr--;
  }
  return bnd;
}
int predfs(int i, int p) {
  int ret = 1;
  for (int a : adj[i]) {
    if (a == p) continue;
    ret += predfs(a, i);
  }
  ccnt[i] = ret;
  return ret;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    cin >> a >> b;
    adj[a].push_back(b);
    adj[b].push_back(a);
  }
  ccnt[1] = predfs(1, 0);
  dfs(1, 0, 1, 2 + (int)adj[1].size());
  for (int i = 1; i <= n; i++)
    cout << ans[i].first << ' ' << ans[i].second << '\n';
}