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;
vector<int> g[500005];
int sz[500005];
pair<int, int> answer[500005];
void count_size(int v, int pr) {
  sz[v] = 1;
  for (int i = 0; i < (int)g[v].size(); i++) {
    int u = g[v][i];
    if (u != pr) {
      count_size(u, v);
      sz[v] += sz[u];
    }
  }
  return;
}
void dfs(int v, int pr, int lef, int rig, bool fl) {
  if (fl) {
    int uk = lef;
    for (int i = 0; i < (int)g[v].size(); i++) {
      int u = g[v][i];
      if (u == pr) continue;
      answer[u].first = uk;
      uk++;
    }
    answer[v].first = uk;
    uk++;
    for (int i = (int)g[v].size() - 1; i >= 0; i--) {
      int u = g[v][i];
      if (u == pr) continue;
      dfs(u, v, uk, uk + 2 * sz[u] - 2, false);
      uk += 2 * sz[u] - 1;
    }
  } else {
    int uk = rig;
    for (int i = 0; i < (int)g[v].size(); i++) {
      int u = g[v][i];
      if (u == pr) continue;
      answer[u].second = uk;
      uk--;
    }
    answer[v].second = uk;
    uk--;
    for (int i = (int)g[v].size() - 1; i >= 0; i--) {
      int u = g[v][i];
      if (u == pr) continue;
      dfs(u, v, uk - 2 * sz[u] + 2, uk, true);
      uk -= 2 * sz[u] - 1;
    }
  }
  return;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  for (int i = 1; i < n; i++) {
    int u, v;
    cin >> u >> v;
    g[u].push_back(v), g[v].push_back(u);
  }
  answer[1].second = 2 * n;
  count_size(1, -1);
  dfs(1, -1, 1, 2 * n - 1, true);
  for (int i = 1; i <= n; i++)
    cout << answer[i].first << ' ' << answer[i].second << '\n';
  return 0;
}