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 long long mod = 1e9 + 7;
const long long inf = 1e9 + 7;
const long long MAX = 500010;
const long double EPS = 1e-9;
vector<vector<long long> > G(MAX);
vector<long long> l(MAX);
vector<long long> r(MAX);
long long now;
void dfs(long long i, long long p) {
  long long sz = G[i].size();
  for (int j = sz - 1; j >= 0; j--) {
    long long e = G[i][j];
    if (e == p) continue;
    l[e] = now;
    now++;
  }
  r[i] = now;
  now++;
  for (int j = (0); j < (sz); j++) {
    long long e = G[i][j];
    if (e == p) continue;
    dfs(e, i);
  }
}
signed main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  long long n;
  cin >> n;
  for (int i = (0); i < (n - 1); i++) {
    long long a, b;
    cin >> a >> b;
    a--;
    b--;
    G[a].push_back(b);
    G[b].push_back(a);
  }
  now = 2;
  l[0] = 1;
  dfs(0, -1);
  for (int i = (0); i < (n); i++) {
    cout << l[i] << ' ' << r[i] << '\n';
  }
}