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 N = (long long)(1e6) + 322;
const long long inf = (long long)(1e18) + 123;
const long long mod = (long long)(1e9) + 7;
const double EPS = 1e-9;
vector<long long> v[N];
long long l[N], r[N], n, val;
long long dfs(long long x, long long p = -1) {
  for (auto to : v[x]) {
    if (to == p) continue;
    dfs(to, x);
  }
  l[x] = ++val;
  reverse((v[x]).begin(), (v[x]).end());
  for (auto to : v[x]) {
    if (to == p) continue;
    r[to] = ++val;
  }
  return val;
}
signed main() {
  ios_base ::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (long long i = 1; i < n; ++i) {
    long long x, y;
    cin >> x >> y;
    v[x].push_back(y);
    v[y].push_back(x);
  }
  r[1] = n * 2;
  dfs(1, 0);
  for (long long i = 1; i <= n; ++i) {
    cout << l[i] << ' ' << r[i] << "\n";
  }
  return 0;
}