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 long long MOD = 1e9 + 7;
vector<int> g[500050];
pair<int, int> ans[500050];
int size[500050];
void setSize(int now, int par) {
  size[now] = 1;
  for (auto nex : g[now]) {
    if (nex == par) continue;
    setSize(nex, now);
    size[now] += size[nex];
  }
}
void dfs(int now, int par) {
  int cen = ans[now].first;
  vector<int> nex;
  for (auto n : g[now]) {
    if (par != n) nex.push_back(n);
  }
  int pos = cen;
  for (int i = 0; i < nex.size(); i++) {
    int tar = nex[i];
    int cnt = pos - (g[tar].size());
    ans[tar].first = cnt;
    ans[tar].second = cen + i + 1;
    dfs(tar, now);
    pos -= g[tar].size();
    pos -= (2 * size[tar] - g[tar].size() - 1);
  }
}
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  int N;
  cin >> N;
  vector<int> L(N - 1), R(N - 1);
  for (int i = 0; i < (N - 1); ++i) cin >> L[i] >> R[i];
  for (int i = 0; i < (N - 1); ++i) {
    L[i]--;
    R[i]--;
    g[L[i]].push_back(R[i]);
    g[R[i]].push_back(L[i]);
  }
  ans[0].first = 2 * N - g[0].size() - 1;
  ans[0].second = 2 * N;
  setSize(0, -1);
  dfs(0, -1);
  for (int i = 0; i < (N); ++i) printf("%d %d\n", ans[i].first, ans[i].second);
  return 0;
}