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;
template <typename T>
void min_self(T& a, T b) {
  a = min(a, b);
}
template <typename T>
void max_self(T& a, T b) {
  a = max(a, b);
}
using ll = long long;
const int MAXN = 500 * 1000 + 5;
int a[2 * MAXN];
vector<vector<int>> g;
vector<int> sz;
vector<pair<int, int>> ans;
void dfs_init(int u, int p) {
  auto it = find(g[u].begin(), g[u].end(), p);
  if (it != g[u].end()) {
    g[u].erase(it);
  }
  sz[u] = 1;
  for (int v : g[u]) {
    if (v == p) {
      continue;
    }
    dfs_init(v, u);
    sz[u] += sz[v];
  }
}
void dfs(int u, int start) {
  int children = g[u].size();
  ans[u].second = a[start + children + 1];
  for (int i = 0; i < int(children); i++) {
    int v = g[u][i];
    ans[v].first = a[start + i + 1];
  }
  int new_start = ans[u].second - 1;
  for (int i = children - 1; i >= 0; i--) {
    int v = g[u][i];
    dfs(v, new_start);
    new_start += 2 * sz[v] - 1;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  g.resize(n);
  ans.resize(n);
  sz.resize(n);
  for (int i = 0; i < int(n - 1); i++) {
    int u, v;
    cin >> u >> v;
    u--, v--;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  for (int i = 0; i < int(2 * n); i++) {
    a[i] = i + 1;
  }
  dfs_init(0, -1);
  ans[0].first = 1;
  dfs(0, 0);
  for (const pair<int, int>& e : ans) {
    cout << e.first << ' ' << e.second << '\n';
  }
}