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 N = (int)5e5 + 9;
vector<int> T[N];
bool use[2 * N];
int p;
int l[N], r[N];
void dfs(int u, int par) {
  if (par == -1) {
    l[u] = 1;
    r[u] = T[u].size() + 2;
    use[1] = true;
    use[T[u].size() + 2] = true;
    p = T[u].size() + 3;
    vector<pair<int, int> > srt;
    int fl = 2;
    for (auto x : T[u]) {
      if (x == par) continue;
      dfs(x, u);
      srt.push_back(make_pair(r[x], x));
    }
    sort(srt.begin(), srt.end());
    reverse(srt.begin(), srt.end());
    for (int i = 0; i < srt.size(); i++) {
      l[srt[i].second] = fl++;
    }
    return;
  }
  vector<int> rr;
  for (auto x : T[u]) {
    if (x == par) {
      continue;
    }
    while (use[p]) p++;
    use[p] = true;
    rr.push_back(p);
  }
  while (use[p]) p++;
  r[u] = p;
  use[p] = true;
  vector<pair<int, int> > srt;
  for (int i = 0; i < T[u].size(); i++) {
    if (T[u][i] == par) continue;
    dfs(T[u][i], u);
    srt.push_back(make_pair(r[T[u][i]], T[u][i]));
  }
  sort(srt.begin(), srt.end());
  reverse(srt.begin(), srt.end());
  for (int i = 0; i < srt.size(); i++) {
    l[srt[i].second] = rr[i];
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  int n;
  cin >> n;
  int x, y;
  for (int i = 1; i < n; i++) {
    cin >> x >> y;
    T[x].push_back(y);
    T[y].push_back(x);
  }
  dfs(1, -1);
  for (int i = 1; i <= n; i++) cout << l[i] << " " << r[i] << "\n";
  return 0;
}