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 N = 5e5 + 5;
struct node {
  int id, type;
  node* nxt = nullptr;
  node(){};
  node(int id, int type) : id(id), type(type){};
};
int n;
node* head[N];
node* tail[N];
vector<int> adj[N];
pair<int, int> res[N];
void dfs(int u, int p) {
  vector<node*> p1, p2, p3;
  for (int v : adj[u]) {
    if (v == p) {
      continue;
    }
    dfs(v, u);
    p1.push_back(head[v]);
    p2.push_back(head[v]->nxt);
    p3.push_back(tail[v]);
  }
  head[u] = new node(u, 0);
  tail[u] = new node(u, 1);
  node* cur = head[u];
  int sz = p1.size();
  for (int i = 0; i < sz; ++i) {
    cur->nxt = p1[i];
    cur = p1[i];
  }
  cur->nxt = tail[u];
  cur = tail[u];
  for (int i = sz - 1; i >= 0; --i) {
    cur->nxt = p2[i];
    cur = p3[i];
  }
  tail[u] = cur;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1, u, v; i < n; ++i) {
    cin >> u >> v;
    adj[u].emplace_back(v);
    adj[v].emplace_back(u);
  }
  dfs(1, -1);
  node* cur = head[1];
  for (int i = 1; i <= 2 * n; ++i) {
    if (cur->type == 0) {
      res[cur->id].first = i;
    } else {
      res[cur->id].second = i;
    }
    cur = cur->nxt;
  }
  for (int i = 1; i <= n; ++i) {
    cout << res[i].first << " " << res[i].second << '\n';
  }
  return 0;
}