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 N = 500 * 1000 + 13;
int n;
vector<int> g[N], vs[N];
pair<int, int> segs[N];
void dfs(int v, int p = -1) {
  int sum = 0;
  int bst = -1;
  for (auto to : g[v])
    if (to != p) {
      dfs(to, v);
      sum += 2 * int((vs[to]).size());
      if (bst == -1 || int((vs[to]).size()) > int((vs[bst]).size())) bst = to;
    }
  if (bst == -1) {
    vs[v].push_back(v);
    segs[v] = make_pair(1, 2);
    return;
  }
  swap(vs[v], vs[bst]);
  int lst = segs[bst].second;
  sum -= 2 * int((vs[v]).size());
  sum += 1;
  segs[bst].second += sum;
  for (auto to : g[v])
    if (to != p && to != bst) {
      int add = lst - 1;
      for (auto u : vs[to]) {
        segs[u].first += add;
        segs[u].second += add;
        vs[v].push_back(u);
      }
      lst = segs[to].second;
      sum -= 2 * int((vs[to]).size());
      segs[to].second += sum;
      vs[to].clear();
    }
  vs[v].push_back(v);
  segs[v] = make_pair(lst, segs[bst].second + 1);
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < int(n - 1); i++) {
    int first, second;
    scanf("%d%d", &first, &second);
    --first;
    --second;
    g[first].push_back(second);
    g[second].push_back(first);
  }
  dfs(0);
  for (int i = 0; i < n; i++) printf("%d %d\n", segs[i].first, segs[i].second);
  return 0;
}