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 = 555555;
vector<int> g[maxn];
int n;
pair<int, int> ans[maxn];
int sz[maxn];
void dfs1(int u, int p) {
  for (auto v : g[u]) {
    if (v == p) continue;
    dfs1(v, u);
    sz[u]++;
  }
}
int dfs(int u, int p, int l, int r) {
  ans[u].first = l;
  ans[u].second = r + sz[u];
  int pre = ans[u].second;
  int now = ans[u].second;
  int mx = ans[u].second;
  for (auto v : g[u]) {
    if (v == p) continue;
    mx = max(dfs(v, u, --now, mx + 1), mx);
  }
  return mx;
}
int main() {
  ios::sync_with_stdio(false);
  int n;
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    cin >> u >> v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  dfs1(1, -1);
  dfs(1, -1, 1, 2);
  for (int i = 1; i <= n; i++) {
    cout << ans[i].first << " " << ans[i].second << endl;
  }
  return 0;
}