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;
using ll = long long;
using ld = long double;
int MOD = 1000000007;
ll INF = 1LL << 62;
ld EPS = 1e-8;
ld PI = acos(-1);
int N;
vector<int> adj[500010];
vector<int> child[500010];
int lft[500010], rgt[500010];
int pos = 0;
void dfs(int p, int pp) {
  for (int i : adj[p]) {
    if (i != pp) {
      child[p].push_back(i);
      dfs(i, p);
    }
  }
}
void dfs1(int p) {
  for (int i : child[p]) lft[i] = pos++;
  rgt[p] = pos++;
  for (int i = (int)child[p].size() - 1; i >= 0; i--) dfs1(child[p][i]);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout << fixed << setprecision(10);
  cin >> N;
  for (int i = 0; i < N - 1; i++) {
    int a, b;
    cin >> a >> b;
    a--, b--;
    adj[a].push_back(b);
    adj[b].push_back(a);
  }
  dfs(0, -1);
  lft[0] = pos++;
  dfs1(0);
  for (int i = 0; i < N; i++) {
    cout << lft[i] + 1 << " " << rgt[i] + 1 << "\n";
  }
  return 0;
}