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 int64_t M1 = 998244353;
const int64_t M2 = 1000000007;
const int64_t inf = 1e18;
mt19937 rng((uint64_t)chrono::steady_clock::now().time_since_epoch().count());
vector<int64_t> l(500001), r(500001);
vector<vector<int64_t>> node(500001);
void dfs(int64_t u, int64_t p, int64_t& m) {
  int64_t x = m + (node[u].size() - (p != -1));
  r[u] = x;
  m = x + 1;
  for (int64_t v : node[u])
    if (v != p) {
      l[v] = --x;
      dfs(v, u, m);
    }
}
void solve() {
  int64_t n;
  cin >> n;
  for (int64_t i = 1; i < n; i++) {
    int64_t u, v;
    cin >> u >> v;
    node[u - 1].push_back(v - 1);
    node[v - 1].push_back(u - 1);
  }
  l[0] = 1;
  int64_t m = 2;
  dfs(0, -1, m);
  for (int64_t i = 0; i < n; i++) cout << l[i] << " " << r[i] << "\n";
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  solve();
}