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 <class T>
ostream& operator<<(ostream& os, vector<T> V) {
  os << "[ ";
  for (auto v : V) os << v << " ";
  return os << "]";
}
template <class T>
ostream& operator<<(ostream& os, set<T> S) {
  os << "{ ";
  for (auto s : S) os << s << " ";
  return os << "}";
}
template <class L, class R>
ostream& operator<<(ostream& os, pair<L, R> P) {
  return os << "(" << P.first << "," << P.second << ")";
}
template <class L, class R>
ostream& operator<<(ostream& os, map<L, R> M) {
  os << "{ ";
  for (auto m : M) os << "(" << m.F << ":" << m.S << ")";
  return os << "}";
}
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  cout << name << " : " << arg1 << std::endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  cout.write(names, comma - names) << " : " << arg1 << " | ";
  __f(comma + 1, args...);
}
const int N = 5e5 + 100;
vector<int> G[N];
int u, v, n;
int L[N], R[N], cnt[N];
void dfs(int s, int p) {
  cnt[s] = 1;
  for (auto it : G[s]) {
    if (it == p) continue;
    dfs(it, s);
    cnt[s] += cnt[it];
  }
}
int solve(int s, int p, int l, int r) {
  L[s] = l, R[s] = r;
  int no = 1;
  int lll = l;
  for (auto it : G[s]) {
    if (it == p) continue;
    int rr = l + no;
    no++;
    lll -= G[it].size();
    lll = solve(it, s, lll, rr);
  }
  return lll;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cout << fixed << setprecision(25);
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    cin >> u >> v;
    G[u].push_back(v);
    G[v].push_back(u);
  }
  dfs(1, -1);
  solve(1, -1, 2 * n - G[1].size() - 1, 2 * n);
  for (int i = 1; i <= n; ++i) {
    assert(L[i] >= 1 && R[i] <= 2 * n);
    cout << L[i] << " " << R[i] << "\n";
  }
  return 0;
}