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 INF = 1000000000;
struct edge {
  int x, y, z;
  edge(int a, int b, int c) {
    x = a;
    y = b;
    z = c;
  }
  bool operator<(edge a) const {
    return vector<int>{x, y, z} < vector<int>{a.x, a.y, a.z};
  }
};
int vp[511000][2];
vector<int> G[510000];
int vi[511000];
int now = 1;
void dfs(int p, int from) {
  if (vi[p]) return;
  vi[p] = 1;
  for (int adj : G[p]) dfs(adj, p);
  vp[p][0] = now++;
  auto &tv = G[p];
  for (int i = tv.size() - 1; i >= 0; i--) {
    int tmp = tv[i];
    if (tmp != from) vp[tmp][1] = now++;
  }
}
int main() {
  std::ios::sync_with_stdio(false);
  std::cin.tie(0);
  int n;
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    cin >> a >> b;
    G[a].push_back(b);
    G[b].push_back(a);
  }
  dfs(1, 0);
  vp[1][1] = now;
  for (int i = 1; i <= n; i++) cout << vp[i][0] << " " << vp[i][1] << endl;
}