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;
long long power(long long x, long long y, long long md = 1000000007) {
  long long res = 1;
  x %= md;
  while (y > 0) {
    if (y & 1) res = (res * x) % md;
    x = (x * x) % md;
    y = y >> 1;
  }
  return res % md;
}
struct node {
  vector<long long> v, w;
  long long sz;
} nodes[500004];
vector<pair<long long, long long> > ans(500005);
long long dfs(long long index, long long parent) {
  long long sz = 1;
  long long len = nodes[index].v.size();
  for (long long i = 0; i < len; i++) {
    if (nodes[index].v[i] != parent) {
      sz += dfs(nodes[index].v[i], index);
    }
  }
  return nodes[index].sz = sz;
}
void dfs1(long long index, long long parent, long long st, long long used) {
  long long sz = 0;
  long long cnt = 0;
  long long len = nodes[index].v.size();
  for (long long i = 0; i < len; i++) {
    if (nodes[index].v[i] != parent) {
      cnt++;
      sz += 2 * nodes[nodes[index].v[i]].sz - 1;
    }
  }
  ans[index] = {st, st + used + cnt + 1};
  long long us = 0;
  long long st1 = st + used + cnt + 1;
  for (long long i = 0; i < len; i++) {
    if (nodes[index].v[i] != parent) {
      st1--;
      dfs1(nodes[index].v[i], index, st1, us + (st + used + cnt + 1 - st1));
      us += 2 * nodes[nodes[index].v[i]].sz - 1;
    }
  }
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  long long n;
  cin >> n;
  for (long long i = 0; i < n - 1; i++) {
    long long u, v;
    cin >> u >> v;
    u--;
    v--;
    nodes[u].v.push_back(v);
    nodes[v].v.push_back(u);
  }
  dfs(0, -1);
  dfs1(0, -1, 1, 0);
  for (long long i = 0; i < n; i++)
    cout << ans[i].first << " " << ans[i].second << '\n';
}