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 gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
bool isPrime(long long n) {
  if (n < 2) return 0;
  if (n < 4) return 1;
  if (n % 2 == 0 or n % 3 == 0) return 0;
  for (long long i = 5; i * i <= n; i += 6)
    if (n % i == 0 or n % (i + 2) == 0) return 0;
  return 1;
}
long long modexpo(long long x, long long p) {
  long long res = 1;
  x = x % 1000000007;
  while (p) {
    if (p % 2) res = res * x;
    p >>= 1;
    x = x * x % 1000000007;
    res %= 1000000007;
  }
  return res;
}
long long vis[2000005];
vector<pair<long long, long long>> v(2000005);
vector<long long> adj[2000005];
long long c = 2;
void dfs(long long u) {
  vis[u] = 1;
  for (long long i = 0; i < adj[u].size(); ++i) {
    if (!vis[adj[u][i]]) {
      v[adj[u][i]].first = c++;
    }
  }
  v[u].second = c++;
  long long n = adj[u].size();
  for (long long i = n - 1; i >= 0; --i) {
    if (!vis[adj[u][i]]) {
      dfs(adj[u][i]);
    }
  }
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  while (t--) {
    long long n;
    cin >> n;
    long long x, y;
    for (long long i = 0; i < n - 1; ++i) {
      cin >> x >> y;
      adj[x].push_back(y);
      adj[y].push_back(x);
    }
    v[1].first = 1;
    dfs(1);
    for (long long i = 0; i < n; ++i) {
      cout << v[i + 1].first << " " << v[i + 1].second << "\n";
    }
    cout << "\n";
  }
}