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 N = 5e5 + 5;
long long n;
vector<long long> pos;
vector<long long> g[N];
void dfs(long long u, long long par) {
  for (auto &it : g[u]) {
    if (it == par) continue;
    pos.push_back(it);
  }
  pos.push_back(u);
  long long sz = g[u].size();
  for (long long i = sz - 1; i >= 0; i--) {
    if (g[u][i] == par) continue;
    dfs(g[u][i], u);
  }
}
vector<long long> ans[N];
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  cin >> n;
  for (long long i = 1; i <= n - 1; i++) {
    long long u, v;
    cin >> u >> v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  pos.push_back(1);
  dfs(1, 1);
  for (long long i = 0; i < pos.size(); i++) ans[pos[i]].push_back(i + 1);
  for (long long i = 1; i <= n; i++) {
    for (auto &it : ans[i]) cout << it << " ";
    cout << "\n";
  }
  return 0;
}