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;
int dx[4] = {1, 0, -1, 0}, dy[4] = {0, 1, 0, -1};
int ddx[8] = {1, 1, 0, -1, -1, -1, 0, 1}, ddy[8] = {0, 1, 1, 1, 0, -1, -1, -1};
void mad(long long &a, long long b) {
  a = (a + b) % 1000000007;
  if (a < 0) a += 1000000007;
}
long long gcd(long long a, long long b) {
  if (!a) return b;
  return gcd(b % a, a);
}
vector<int> ed[500005];
vector<int> timeline, intervals[500005];
void dfs(int i, int p) {
  for (int j : ed[i]) {
    if (j == p) continue;
    dfs(j, i);
  }
  timeline.push_back(i);
  reverse(ed[i].begin(), ed[i].end());
  for (int j : ed[i])
    if (j != p) timeline.push_back(j);
}
void solve() {
  int n;
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    cin >> u >> v;
    ed[u].push_back(v);
    ed[v].push_back(u);
  }
  dfs(1, 0);
  timeline.push_back(1);
  for (int i = 0; i < timeline.size(); i++) {
    intervals[timeline[i]].push_back(i + 1);
  }
  for (int i = 1; i <= n; i++)
    cout << intervals[i][0] << " " << intervals[i][1] << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t;
  t = 1;
  while (t--) {
    solve();
  }
}