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 maxN = 600000;
int N, a, b;
vector<int> graph[maxN];
bool checked[maxN];
int L[maxN], R[maxN], nowright[maxN], nowleft[maxN];
int dfs(int p, int last_end) {
  checked[p] = true;
  int s = graph[p].size() + last_end;
  nowright[p] = s;
  R[p] = s;
  nowleft[p] = s;
  for (int i = 0; i < graph[p].size(); i++) {
    int np = graph[p][i];
    if (!checked[np]) {
      nowright[p] -= 1;
      L[np] = nowright[p];
      nowleft[p] = dfs(np, nowleft[p]);
    }
  }
  return nowleft[p];
}
int main() {
  cin >> N;
  for (int i = 0; i < N - 1; i++) {
    cin >> a >> b;
    graph[a - 1].push_back(b - 1);
    graph[b - 1].push_back(a - 1);
  }
  L[0] = 1;
  dfs(0, 2);
  for (int i = 0; i < N; i++) {
    cout << L[i] << " " << R[i] << "\n";
  }
}