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;
int const oo = 1000000007, e5 = 100007, e6 = 1000007;
int N;
vector<int> edge[5 * e5];
stack<int> dissolve;
int answer[e6];
int currWrite = 1;
bool ck[5 * e5];
int startOf[5 * e5], endOf[5 * e5];
int __init__() {
  for (int i = 0; i < 500005; i++) {
    ck[i] = 0;
    startOf[i] = 0;
    endOf[i] = 0;
  }
  return 0;
}
void dissolveNode(int node) {
  int child;
  ck[node] = 1;
  for (int i = 0; i < edge[node].size(); i++) {
    child = edge[node][i];
    if (ck[child] == 1) continue;
    answer[currWrite] = child;
    ++currWrite;
    dissolve.push(child);
  }
  answer[currWrite] = node;
  ++currWrite;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  __init__();
  int a, b;
  cin >> N;
  for (int i = 0; i < N - 1; i++) {
    cin >> a >> b;
    edge[a].push_back(b);
    edge[b].push_back(a);
  }
  answer[1] = 1;
  currWrite = 2;
  dissolve.push(1);
  while (!dissolve.empty()) {
    int next = dissolve.top();
    dissolve.pop();
    dissolveNode(next);
  }
  for (int i = 1; i <= 2 * N; i++) {
    if (startOf[answer[i]] == 0) {
      startOf[answer[i]] = i;
    } else {
      endOf[answer[i]] = i;
    }
  }
  for (int i = 1; i <= N; i++) {
    cout << startOf[i] << " " << endOf[i] << "\n";
  }
}