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>
#pragma GCC optimize("O3")
using namespace std;
const long long N = 1e6 + 10;
const long long inf = 1e18;
const long long mod = 1e9 + 7;
const long long eps = 1e-9;
const char spc = 29;
mt19937 rnd(chrono::high_resolution_clock::now().time_since_epoch().count());
using namespace std;
long long n;
vector<long long> g[N];
vector<long long> order;
void dfs(long long v, long long p) {
  vector<long long> nxt;
  for (auto u : g[v])
    if (u != p) {
      order.push_back(u);
      nxt.push_back(u);
    }
  order.push_back(v);
  reverse(nxt.begin(), nxt.end());
  for (auto u : nxt) dfs(u, v);
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (long long i = 0; i < n - 1; ++i) {
    long long v, u;
    cin >> v >> u;
    --v;
    --u;
    g[v].push_back(u);
    g[u].push_back(v);
  }
  order.push_back(0);
  dfs(0, 0);
  vector<pair<long long, long long> > ans(n);
  set<long long> st;
  long long cur = 1;
  for (auto it : order)
    if (st.find(it) == st.end()) {
      st.insert(it);
      ans[it].first = cur++;
    } else
      ans[it].second = cur++;
  for (auto &p : ans) cout << p.first << " " << p.second << "\n";
}