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;
using ll = long long;
const int maxn = 5e5 + 5;
int n;
vector<int> adj[maxn];
int len[maxn];
int sz[maxn];
pair<int, int> ans[maxn];
void dfs(int i, int p) {
  len[i] = 2;
  sz[i] = 2;
  int add = 1;
  for (int j : adj[i]) {
    if (j == p) continue;
    dfs(j, i);
    len[i] += len[j];
    sz[i]++;
    len[j] += add;
    sz[j] += add;
    add = len[j];
  }
}
void recover(int i, int p) {
  int off = -1;
  for (int j : adj[i]) {
    if (j == p) continue;
    ans[j].first = ans[i].second + off;
    off--;
    ans[j].second = ans[j].first + sz[j] - 1;
    recover(j, i);
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    int x, y;
    cin >> x >> y;
    adj[x].push_back(y);
    adj[y].push_back(x);
  }
  dfs(1, 1);
  ans[1] = {1, sz[1]};
  recover(1, 1);
  for (int i = 1; i <= n; i++) {
    cout << ans[i].first << ' ' << ans[i].second << '\n';
  }
}