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;
int n, x, y, i, siz[500005], seq[1000005], e[2][500005];
vector<int> s[500005];
void dfs(int p, int f) {
  int i, v;
  siz[p] = 1;
  for (i = 0; i < s[p].size(); i++) {
    v = s[p][i];
    if (v == f) continue;
    dfs(v, p);
    siz[p] += siz[v];
  }
}
void dfs2(int l, int r, int p, int f) {
  if (l == r) {
    seq[l] = p;
    return;
  }
  int i, v, pre;
  pre = l - 1;
  for (i = 0; i < s[p].size(); i++) {
    v = s[p][i];
    if (v == f) continue;
    dfs2(pre + 1, pre + 2 * siz[v] - 1, v, p);
    pre += 2 * siz[v] - 1;
  }
  seq[++pre] = p;
  for (i = s[p].size() - 1; i >= 0; i--) {
    v = s[p][i];
    if (v == f) continue;
    seq[++pre] = v;
  }
}
int main() {
  scanf("%d", &n);
  for (i = 1; i < n; i++) {
    scanf("%d%d", &x, &y);
    s[x].push_back(y);
    s[y].push_back(x);
  }
  dfs(1, 0);
  seq[2 * n] = 1;
  dfs2(1, 2 * n - 1, 1, 0);
  for (i = 1; i <= 2 * n; i++) {
    if (e[0][seq[i]] == 0)
      e[0][seq[i]] = i;
    else
      e[1][seq[i]] = i;
  }
  for (i = 1; i <= n; i++) printf("%d %d\n", e[0][i], e[1][i]);
  return 0;
}