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;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
using vi = vector<int>;
using vl = vector<ll>;
const int MAXN = 5e5 + 5;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
vi g[MAXN];
int at = 2;
int mar[MAXN];
int first_livre = 2;
void dfs(int u, int p, vector<pii> &a) {
  if (mar[u] == 1) return;
  mar[u] = 1;
  int sz = g[u].size();
  if (p != -1) sz -= 1;
  a[u].second = at + sz;
  at = a[u].second + 1;
  if (p == -1) {
    a[u].first = 1;
  }
  int x = 1;
  for (int v : g[u]) {
    if (mar[v] == 0) a[v].first = a[u].second - x;
    if (v == p) continue;
    dfs(v, u, a);
    x++;
  }
}
void solve() {
  int n;
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    cin >> u >> v;
    u--;
    v--;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  vector<pii> a(n);
  dfs(0, -1, a);
  for (int i = 0; i < n; i++) {
    cout << a[i].first << " " << a[i].second << '\n';
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  int t = 1;
  while (t--) {
    solve();
  }
  return 0;
}