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 INF = 0x3f3f3f3f;
int all[500000 + 10], l[500000 + 10], used[1000000 + 10], r[500000 + 10], n,
    children[500000 + 10], ok;
vector<int> each[500000 + 10];
void dfs(int now, int prev) {
  if (prev == -1) {
    children[now] = each[now].size();
  } else {
    children[now] = each[now].size() - 1;
  }
  for (auto it : each[now]) {
    if (it != prev) {
      dfs(it, now);
      all[now] += all[it];
    }
  }
  all[now] += children[now];
}
void build(int now, int prev) {
  int tmp = ok;
  l[now] = ok + all[now] * 2 - children[now] + 1;
  for (auto it : each[now]) {
    if (it != prev) {
      build(it, now);
      ok--;
    }
  }
  ok = tmp + (all[now] + 1) * 2;
}
vector<int> r_bound;
int pt = 0;
void calc(int now, int prev) {
  for (auto it : each[now]) {
    if (it != prev) {
      calc(it, now);
    }
  }
  int SZ = each[now].size();
  if (SZ)
    for (int i = SZ - 1; i >= 0; --i)
      if (each[now][i] != prev) r[each[now][i]] = r_bound[pt++];
}
int main() {
  cin >> n;
  for (int i = 1; i <= n - 1; ++i) {
    int u, v;
    scanf("%d %d", &u, &v);
    each[u].push_back(v);
    each[v].push_back(u);
  }
  dfs(1, -1);
  build(1, -1);
  for (int i = 1; i <= n; ++i) used[l[i]] = 1;
  for (int i = 1; i <= 2 * n; ++i) {
    if (!used[i]) {
      r_bound.push_back(i);
    }
  }
  sort(r_bound.begin(), r_bound.end());
  calc(1, -1);
  r[1] = r_bound[pt++];
  for (int i = 1; i <= n; ++i) {
    printf("%d %d\n", l[i], r[i]);
  }
  return 0;
}