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 n, u, v, y;
vector<long long> adj[500005];
bool vis[500005];
pair<long long, long long> ad[500005];
void dfs(long long x) {
  if (vis[x]) return;
  vis[x] = 1;
  if (x != 1)
    ad[x].second = y + adj[x].size() - 1;
  else
    ad[x].second = y + adj[x].size();
  for (long long i = 0; i < adj[x].size(); i++) {
    if (vis[adj[x][i]]) continue;
    ad[adj[x][i]].first = y;
    y++;
  }
  y++;
  for (long long i = adj[x].size() - 1; i >= 0; i--) dfs(adj[x][i]);
}
int main() {
  scanf("%lld", &n);
  if (n == 1) {
    cout << "1 2";
    return 0;
  }
  for (long long i = 0; i < n - 1; i++) {
    scanf("%lld%lld", &u, &v);
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  ad[1].first = 1;
  y = 2;
  dfs(1);
  for (long long i = 1; i <= n; i++)
    printf("%lld %lld\n", ad[i].first, ad[i].second);
}