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<vector<long long>> g;
vector<long long> ls, rs, used;
long long from;
void dfs(long long pos, long long par) {
  long long children = g[pos].size() - (par == -1 ? 0 : 1);
  rs[pos] = from + children;
  long long lp = rs[pos] - 1;
  from = rs[pos] + 1;
  for (int i = 0; i < g[pos].size(); ++i) {
    long long nxt = g[pos][i];
    if (nxt == par) {
      continue;
    }
    ls[nxt] = lp--;
    dfs(nxt, pos);
  }
}
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  long long n;
  cin >> n;
  g = vector<vector<long long>>(n);
  ls = vector<long long>(n);
  rs = vector<long long>(n);
  used = vector<long long>(2 * n + 2, 0);
  for (int i = 0; i < n - 1; ++i) {
    long long fr, to;
    cin >> fr >> to;
    fr--;
    to--;
    g[fr].push_back(to);
    g[to].push_back(fr);
  }
  ls[0] = 1;
  from = 2;
  dfs(0, -1);
  for (int i = 0; i < n; ++i) {
    cout << ls[i] << " " << rs[i] << "\n";
  }
  return 0;
}