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 maxn = 5e5 + 17;
list<int> ls;
vector<int> out[maxn];
int n;
vector<int> g[maxn];
void dfs(int v = 0, int p = 0, list<int>::iterator w = ls.begin()) {
  for (auto u : g[v])
    if (u != p) {
      ls.insert(next(w), u);
      dfs(u, v, ls.insert(w, u));
    }
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0);
  cin >> n;
  for (int i = 1; i < n; i++) {
    int v, u;
    cin >> v >> u;
    v--, u--;
    g[v].push_back(u);
    g[u].push_back(v);
  }
  ls = {0, 0};
  dfs();
  for (int i = 0; i < 2 * n; i++) {
    static auto it = ls.begin();
    out[*it++].push_back(i);
  }
  for (int i = 0; i < n; i++)
    cout << out[i][0] + 1 << ' ' << out[i][1] + 1 << '\n';
}