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 INF = 0x3f3f3f3f;
const long long LINF = 0x3f3f3f3f3f3f3f3fLL;
const double EPS = 1e-8;
const int MOD = 1000000007;
const int dy[] = {1, 0, -1, 0}, dx[] = {0, -1, 0, 1};
struct IOSetup {
  IOSetup() {
    cin.tie(nullptr);
    ios_base::sync_with_stdio(false);
    cout << fixed << setprecision(20);
    cerr << fixed << setprecision(10);
  }
} iosetup;
int main() {
  int n;
  cin >> n;
  vector<vector<int> > graph(n);
  for (int _ = (0); _ < (n - 1); ++_) {
    int x, y;
    cin >> x >> y;
    --x;
    --y;
    graph[x].emplace_back(y);
    graph[y].emplace_back(x);
  }
  vector<int> l(n, -1), r(n);
  int idx = 0;
  l[0] = idx++;
  vector<int> st;
  st.emplace_back(0);
  while (!st.empty()) {
    int now = st.back();
    st.pop_back();
    sort((graph[now]).begin(), (graph[now]).end());
    for (int e : graph[now]) {
      if (l[e] != -1) continue;
      l[e] = idx++;
      st.emplace_back(e);
    }
    r[now] = idx++;
  }
  for (int i = (0); i < (n); ++i) cout << l[i] + 1 << ' ' << r[i] + 1 << '\n';
  return 0;
}