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];
vector<pair<int, int> > ans(500005);
int i = 1;
void dfs(int cur, int bap) {
  auto it = adj[cur].end();
  it--;
  while (true) {
    if (it == adj[cur].begin()) {
      if (*it != bap) {
        ans[*it].first = i++;
      }
      break;
    }
    if (*it != bap) {
      ans[*it].first = i++;
    }
    it--;
  }
  ans[cur].second = i++;
  for (auto u : adj[cur]) {
    if (u != bap) {
      dfs(u, cur);
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  if (n == 1) {
    cout << "1 2\n";
    return 0;
  }
  for (int i = 0; i < n - 1; i++) {
    int p, q;
    cin >> p >> q;
    adj[p].push_back(q);
    adj[q].push_back(p);
  }
  ans[1].first = i++;
  dfs(1, -1);
  for (int i = 1; i <= n; i++) {
    cout << ans[i].first << " " << ans[i].second << "\n";
  }
}