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 N = 5e5 + 100;
vector<int> g[N], pp(N, -1), a;
void dfs(int v, int p = -1) {
  a.push_back(v);
  for (int to : g[v]) {
    if (to != p) {
      pp[to] = v;
      dfs(to, v);
    }
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
  int n;
  cin >> n;
  for (int i = 1; i < n; i++) {
    int x, y;
    cin >> x >> y;
    g[--x].push_back(--y);
    g[y].push_back(x);
  }
  dfs(0);
  reverse((a).begin(), (a).end());
  vector<int> b;
  for (int i : a) {
    b.push_back(i);
    for (int to : g[i]) {
      if (to != pp[i]) {
        b.push_back(to);
      }
    }
  }
  vector<pair<int, int>> ans(n, {-1, -1});
  for (int i = 0; i < (int)(b).size(); i++) {
    if (ans[b[i]].first == -1) {
      ans[b[i]].first = i + 1;
    } else {
      ans[b[i]].second = i + 1;
    }
  }
  ans[0].second = 2 * n;
  for (int i = 0; i < n; i++) {
    cout << ans[i].first << ' ' << ans[i].second << '\n';
  }
  return 0;
}