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;
long long int gcd(long long int a, long long int b) {
  if (a % b == 0) return b;
  return gcd(b, a % b);
}
long long int poww(long long int a, long long int n) {
  if (n == 0) return 1;
  long long int p = poww(a, n / 2);
  p = (p * p) % 1000000007;
  if (n % 2) p = (a * p) % 1000000007;
  return p;
}
int const nmax = 500006;
vector<int> v[nmax];
int sub[nmax];
long long int ans[nmax][2];
int dfs(int i, int p) {
  sub[i] = 1;
  for (int j = 0; j < v[i].size(); j++) {
    int node = v[i][j];
    if (node != p) {
      sub[i] += dfs(node, i);
    }
  }
  return sub[i];
}
void dfs2(int i, int p, int ind) {
  ans[i][1] = ind;
  long long int cur = 1;
  for (int j = 0; j < v[i].size(); j++) {
    int node = v[i][j];
    if (node != p) {
      ind--;
      ans[node][0] = ind;
      dfs2(node, i, ans[node][0] + cur + v[node].size());
      cur += (2 * sub[node]);
    }
  }
}
void solve() {
  int n;
  cin >> n;
  for (long long int i = 0; i < n - 1; ++i) {
    int x, y;
    cin >> x >> y;
    v[x].push_back(y);
    v[y].push_back(x);
  }
  int num = dfs(1, 0);
  ans[1][0] = 1;
  dfs2(1, 0, ans[1][0] + v[1].size() + 1);
  for (long long int i = 1; i < n + 1; ++i) {
    cout << ans[i][0] << " " << ans[i][1] << endl;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int t = 1;
  while (t--) solve();
  return 0;
}