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;
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
const long long int mod = 1000000007;
const long long int inf = 0x3f3f3f3f3f3f3f3fLL;
const double eps = 1e-9;
const double PI = acos(-1);
long long int i, j, k;
void solve(void);
inline bool equal(double a, double b) { return fabs(a - b) < eps; }
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout << fixed << setprecision(10);
  int t = 1;
  while (t--) solve();
  ;
  ;
  return 0;
}
vector<long long int> node[500005];
bool vis[500005], blocked[1000005];
pair<long long int, long long int> ans[500005];
long long int maxr;
void dfs(int idx, int l) {
  vis[idx] = 1;
  int child = (int)node[idx].size() - 1;
  if (idx == 1) ++child;
  ans[idx].first = l;
  blocked[l] = 1;
  i = maxr + child;
  ans[idx].second = i;
  blocked[i] = 1;
  maxr = ((maxr) > (i + 1) ? (maxr) : (i + 1));
  long long int r = ans[idx].second - 1;
  for (auto it : node[idx])
    if (!vis[it]) {
      dfs(it, r);
      --r;
    }
}
void solve() {
  int n;
  cin >> n;
  for (i = 1; i <= n - 1; ++i) {
    int x, y;
    cin >> x >> y;
    node[x].emplace_back(y);
    node[y].emplace_back(x);
  }
  maxr = 2;
  dfs(1, 1);
  for (i = 1; i <= n; ++i) cout << ans[i].first << ' ' << ans[i].second << '\n';
}