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 mod = 1e9 + 7;
const long long inf = 1e18;
const long double ld_pi = 4 * atan((long double)1);
vector<int> adj_list[1000005];
int ans[1000005][2];
int a[1000005];
bool visited[1000005];
int build(int u, int s, int e) {
  a[s] = u;
  visited[u] = true;
  vector<int> tmp;
  for (int i = 0; i < adj_list[u].size(); i++) {
    int v = adj_list[u][i];
    if (!visited[v]) tmp.push_back(v);
  }
  int last = e + tmp.size() + 1, k = 0;
  a[last - 1] = u;
  for (int i = e + tmp.size() - 1; i >= e; i--, k++) {
    last = build(tmp[k], i, last);
    ++last;
  }
  return last - 1;
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  int n;
  cin >> n;
  for (int i = 1; i < n; i++) {
    int u, v;
    cin >> u >> v;
    adj_list[u].push_back(v);
    adj_list[v].push_back(u);
  }
  build(1, 1, 2);
  for (int i = 1; i <= 2 * n; i++) {
    if (ans[a[i]][0] == 0)
      ans[a[i]][0] = i;
    else
      ans[a[i]][1] = i;
  }
  for (int i = 1; i <= n; i++) {
    cout << ans[i][0] << ' ' << ans[i][1] << endl;
  }
}