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<vector<int> > adj;
vector<int> sz, parent;
vector<pair<int, int> > ret;
int dfs_setup(int curr, int par) {
  int sz_curr = 1;
  parent[curr] = par;
  for (int i = 0; i < adj[curr].size(); ++i) {
    if (adj[curr][i] == par) continue;
    sz_curr += dfs_setup(adj[curr][i], curr);
  }
  sz[curr] = sz_curr;
  return sz_curr;
}
int dfs_solve(int curr, int left, int ch_gap) {
  int num_dir_children = adj[curr].size() - (parent[curr] == -1 ? 0 : 1);
  ret[curr].first = left;
  ret[curr].second = left + ch_gap + num_dir_children + 1;
  int curr_ch_gap = 0;
  int seen = 0;
  for (int i = 0; i < adj[curr].size(); ++i) {
    if (adj[curr][i] == parent[curr]) continue;
    curr_ch_gap +=
        dfs_solve(adj[curr][i], ret[curr].second - seen - 1, curr_ch_gap + 1);
    seen++;
  }
  return 2 * sz[curr];
}
int main() {
  cin.tie(0);
  cin.sync_with_stdio(0);
  int n;
  cin >> n;
  adj.resize(n);
  sz.resize(n);
  parent.resize(n);
  ret.resize(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);
  }
  dfs_setup(0, -1);
  dfs_solve(0, 1, 0);
  for (int i = 0; i < n; ++i) {
    pair<int, int> ret_curr = ret[i];
    cout << ret_curr.first << " " << ret_curr.second << endl;
  }
}