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;
using ULL = unsigned long long;
using PII = pair<int, int>;
using PIL = pair<int, LL>;
using PLL = pair<LL, LL>;
const int mod = 1e9 + 7;
inline void quickread() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
}
int n, x, y;
vector<int> edges[500010];
int l[500010], r[500010];
bool used[500010];
void dfs(int u, int& valid) {
  used[u] = true;
  for (int i = edges[u].size() - 1; i >= 0; --i) {
    int v = edges[u][i];
    if (!used[v]) {
      l[v] = valid++;
    }
  }
  r[u] = valid++;
  for (int i = 0; i < edges[u].size(); ++i) {
    int v = edges[u][i];
    if (!used[v]) {
      dfs(v, valid);
    }
  }
}
int main() {
  quickread();
  cin >> n;
  for (int i = 1; i < n; ++i) {
    cin >> x >> y;
    edges[x].push_back(y);
    edges[y].push_back(x);
  }
  l[1] = 1;
  int valid = 2;
  dfs(1, valid);
  for (int i = 1; i <= n; ++i) {
    cout << l[i] << " " << r[i] << "\n";
  }
  return 0;
}