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 NN = 1 << 20;
int n;
vector<int> adj[NN];
int q[NN];
int par[NN];
int L[NN];
int R[NN];
int cur_num;
void get() {
  memset(par, -1, sizeof(par));
  int beg = 0;
  int fin = 0;
  q[fin++] = 1;
  R[1] = cur_num--;
  par[1] = -2;
  while (beg < fin) {
    int cur = q[beg++];
    for (auto& neighbor : adj[cur]) {
      if (par[neighbor] == -1) {
        par[neighbor] = cur;
        R[neighbor] = cur_num--;
        q[fin++] = neighbor;
      }
    }
    L[cur] = cur_num--;
  }
}
void dfs(int node, int father) {
  for (auto& neighbor : adj[node]) {
    if (neighbor == father) continue;
    R[neighbor] = cur_num--;
  }
  L[node] = cur_num--;
  reverse(adj[node].begin(), adj[node].end());
  for (auto& neighbor : adj[node]) {
    if (neighbor == father) continue;
    dfs(neighbor, node);
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  int x, y;
  for (int i = 0; i < n - 1; i++) {
    cin >> x >> y;
    adj[x].push_back(y);
    adj[y].push_back(x);
  }
  cur_num = 2 * n;
  R[1] = cur_num--;
  dfs(1, -2);
  for (int i = 1; i <= n; i++) {
    cout << L[i] << ' ' << R[i] << '\n';
  }
  return 0;
}