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;
template <typename T = int>
struct Compress {
  map<T, int> id;
  vector<T> num;
  Compress(vector<T> v) {
    sort(v.begin(), v.end());
    v.erase(unique(v.begin(), v.end()), v.end());
    num = v;
    for (int i = 0; i < (int)num.size(); i++) {
      id[num[i]] = i;
    }
  }
  inline int get_id(T x) {
    assert(id.count(x));
    return id[x];
  }
  inline T get_num(int i) {
    assert(0 <= i && i < num.size());
    return num[i];
  }
};
void solve() {
  int n;
  cin >> n;
  vector<vector<int>> g(n);
  for (int _ = 1; _ < n; _++) {
    int x, y;
    cin >> x >> y;
    x--;
    y--;
    g[x].emplace_back(y);
    g[y].emplace_back(x);
  }
  vector<pair<ll, ll>> a(n);
  ll D = 1e7;
  ll l = 0;
  function<void(int, int)> dfs = [&](int u, int p) {
    int i = 1;
    for (int v : g[u])
      if (v != p) {
        a[v] = {l -= D, a[u].first + (i++)};
        dfs(v, u);
      }
  };
  a[0] = {l, D};
  dfs(0, -1);
  vector<ll> b;
  for (int i = 0; i < n; i++) {
    b.push_back(a[i].first);
    b.push_back(a[i].second);
  }
  Compress<ll> c(b);
  for (int i = 0; i < n; i++) {
    int x = c.get_id(a[i].first) + 1;
    int y = c.get_id(a[i].second) + 1;
    cout << x << ' ' << y << "\n";
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  solve();
  cout << endl;
}