1278_E. Tests for problem D_32163
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
#include <bits/stdc++.h>
using namespace std;
const long long MAXN = 500005;
vector<long long> g[MAXN];
long long ind;
pair<long long, long long> ans[MAXN];
void dfs(long long v, long long p) {
  for (long long i = (long long)g[v].size() - 1; i >= 0; i--) {
    if (g[v][i] != p) {
      ans[g[v][i]].second = ind--;
    }
  }
  ans[v].first = ind--;
  for (long long to : g[v]) {
    if (to != p) {
      dfs(to, v);
    }
  }
}
signed main() {
  cin.tie(0);
  cout.tie(0);
  ios_base ::sync_with_stdio(0);
  long long n;
  cin >> n;
  ind = n * 2;
  for (long long i = 0; i < n - 1; i++) {
    long long x, y;
    cin >> x >> y;
    x--;
    y--;
    g[x].push_back(y);
    g[y].push_back(x);
  }
  ans[0].second = ind--;
  dfs(0, -1);
  for (long long i = 0; i < n; i++) {
    cout << ans[i].first << ' ' << ans[i].second << '\n';
  }
  return 0;
}
