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;
mt19937 rnd(chrono::system_clock::now().time_since_epoch().count());
vector<int> vx, vl, vr;
vector<vector<int>> g;
void dfs(int i, int p = -1) {
  for (auto it = g[i].begin(); it != g[i].end(); it++) {
    if (*it != p) {
      dfs(*it, i);
    }
  }
  vl[i] = vx.size();
  vx.push_back(i);
  for (auto it = g[i].rbegin(); it != g[i].rend(); it++) {
    if (*it != p) {
      vr[*it] = vx.size();
      vx.push_back(*it);
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int ii, tt, i, j, k, m, n, c = 0, cc = 0, d, x, y, xl, xr, ans;
  cin >> n;
  vl.resize(n);
  vr.resize(n);
  g.resize(n);
  for (i = 1; i < n; i++) {
    cin >> x >> y;
    x--;
    y--;
    g[x].push_back(y);
    g[y].push_back(x);
  }
  dfs(0);
  vr[0] = vx.size();
  vx.push_back(0);
  for (i = 0; i < n; i++) {
    cout << vl[i] + 1 << ' ' << vr[i] + 1 << '\n';
  }
  return 0;
}