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;
template <typename T>
inline bool chmax(T& a, T b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}
template <typename T>
inline bool chmin(T& a, T b) {
  if (a > b) {
    a = b;
    return true;
  }
  return false;
}
const double EPS = 1e-10;
const double PI = acos(-1);
class Task {
 public:
  vector<vector<int>> G;
  vector<pair<int, int>> seg;
  int dfs1(int s, int p) {
    int r = 0;
    seg[s].first = 1;
    seg[s].second = (int)(G[s]).size() + (p == -1) + 1;
    for (auto u : G[s]) {
      if (u == p) continue;
      int dr = dfs1(u, s);
      seg[u].second += r;
      r += dr;
    }
    return r + 2;
  }
  int dfs2(int s, int p, int left, bool flag) {
    if ((int)(G[s]).size() == 1 and p != -1) return seg[s].second;
    int offset = (int)(G[s]).size() - flag;
    int nleft = left + offset;
    bool nflag = false;
    for (auto u : G[s]) {
      if (u == p) continue;
      seg[u].first = left + offset;
      seg[u].second += left + offset;
      nleft = dfs2(u, s, nleft, nflag);
      --offset;
      nflag = true;
    }
    return nleft;
  }
  void solve(istream& in, ostream& out) {
    int N;
    in >> N;
    G.resize(N);
    seg.resize(N);
    for (int i = (0); i < (N - 1); ++i) {
      int u, v;
      in >> u >> v;
      u--, v--;
      G[u].push_back(v), G[v].push_back(u);
    }
    dfs1(0, -1);
    dfs2(0, -1, 1, false);
    for (auto p : seg) out << p.first << ' ' << p.second << '\n';
  }
};
int main() {
  Task solver;
  istream& in(cin);
  ostream& out(cout);
  ios::sync_with_stdio(false);
  in.tie(nullptr), out.tie(nullptr);
  solver.solve(in, out);
  return 0;
}