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;
const int mod1 = 998244353, mod = 1e9 + 7;
const int MAXN = 5e5 + 6, MAXM = 2e5 + 5;
const int inf = 2e9;
const long long linf = 1e18;
int pos[2 * MAXN], n, x = 0;
vector<int> adj[MAXN], ans[MAXN];
void dfs(int u, int p) {
  for (int v : adj[u])
    if (p != v) pos[++x] = v;
  pos[++x] = u;
  reverse(adj[u].begin(), adj[u].end());
  for (int v : adj[u]) {
    if (p != v) dfs(v, u);
  }
}
void Solve() {
  cin >> n;
  for (int i = 1, u, v; i < n; i++) {
    cin >> u >> v;
    adj[u].emplace_back(v);
    adj[v].emplace_back(u);
  }
  pos[++x] = 1;
  dfs(1, -1);
  for (int i = 1; i <= 2 * n; i++) {
    ans[pos[i]].emplace_back(i);
  }
  cout << '\n';
  for (int i = 1; i <= n; i++) {
    for (auto v : ans[i]) cout << v << ' ';
    cout << '\n';
  }
}
signed main() {
  if (fopen(""
            ".inp",
            "r")) {
    freopen(
        ""
        ".inp",
        "r", stdin);
    freopen(
        ""
        ".out",
        "w", stdout);
  };
  std::ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  int TC = 1;
  while (TC--) Solve();
  return 0;
}