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;
vector<int> adj[500005];
pair<int, int> ans[500005];
int dfs(int st, int pr, int lu, int bg) {
  int ed = lu + adj[st].size();
  ans[st] = {bg, ed};
  int cr = ed - 1;
  int ret = ed;
  for (int x : adj[st]) {
    if (x != pr) {
      ret = dfs(x, st, ret, cr);
      cr--;
    }
  }
  return ret;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n;
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    int x, y;
    cin >> x >> y;
    x--, y--;
    adj[x].push_back(y), adj[y].push_back(x);
  }
  if (n == 1) {
    cout << "1 2\n";
    return 0;
  }
  for (int i = 0; i < n; i++) {
    if (adj[i].size() == 1) {
      ans[i] = {1, 3};
      dfs(adj[i][0], i, 3, 2);
      break;
    }
  }
  for (int i = 0; i < n; i++) {
    cout << ans[i].first << " " << ans[i].second << "\n";
  }
  return 0;
}