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 MAX_N = 500005;
vector<int> G[MAX_N];
int n, k = 1, L[MAX_N], P[MAX_N];
bool odw[MAX_N];
int A[2 * MAX_N];
void dfs(int v) {
  odw[v] = true;
  int d = G[v].size();
  for (int i = d - 1; i >= 0; i--) {
    int u = G[v][i];
    if (odw[u]) continue;
    k++;
    A[k] = u;
  }
  k++;
  A[k] = v;
  for (int i = 0; i < d; i++) {
    int u = G[v][i];
    if (odw[u]) continue;
    dfs(u);
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin >> n;
  for (int i = 1; i < n; i++) {
    int a, b;
    cin >> a >> b;
    G[a].push_back(b);
    G[b].push_back(a);
  }
  A[1] = 1;
  dfs(1);
  for (int i = 1; i <= 2 * n; i++) {
    int x = A[i];
    if (L[x] == 0)
      L[x] = i;
    else
      P[x] = i;
  }
  for (int i = 1; i <= n; i++) cout << L[i] << " " << P[i] << endl;
  return 0;
}