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 N = 5e5 + 5;
bool mark[N];
vector<int> e[N];
pair<int, int> rep[N];
int lf;
void dfs(int n, int l, int r) {
  mark[n] = 1;
  rep[n] = make_pair(l, r);
  r = l;
  for (int y : e[n]) {
    if (mark[y]) continue;
    int cnt = 0;
    for (int z : e[y]) cnt += !mark[z];
    lf -= cnt + 1;
    dfs(y, lf, ++r);
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  int n;
  cin >> n;
  for (int i = 0; i < n - 1; ++i) {
    int x, y;
    cin >> x >> y;
    e[x].push_back(y);
    e[y].push_back(x);
  }
  dfs(1, lf = N, N + e[1].size() + 1);
  int dd = N;
  for (int i = 1; i <= n; ++i) dd = min(dd, rep[i].first);
  for (int i = 1; i <= n; ++i) {
    rep[i].first -= dd - 1;
    rep[i].second -= dd - 1;
  }
  for (int i = 1; i <= n; ++i)
    cout << rep[i].first << " " << rep[i].second << endl;
  return 0;
}