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;
vector<int> v[500010];
int l[500010], r[500010], lp[500010], ml[500010], mr[500010];
int ansl[500010], ansr[500010];
void search(int curr, int prev) {
  int cnt = 0;
  for (auto &to : v[curr]) {
    if (to == prev) continue;
    if (cnt++ == 0) {
      lp[curr] = to;
      search(to, curr);
      l[curr] += l[to] + ml[to];
      ml[curr] += r[to] + mr[to];
    } else {
      search(to, curr);
      r[curr] += r[to] + mr[to];
      mr[curr] += l[to] + ml[to];
    }
  }
  l[curr]++;
  r[curr]++;
}
void search2(int curr, int prev, int li, int ri) {
  {
    ansl[curr] = l[curr] + li - 1;
    ansr[curr] = ri - r[curr] + 1;
    int lsum = 0, rsum = 0;
    for (auto &to : v[curr]) {
      if (to == prev) continue;
      if (lp[curr] == to) {
        search2(to, curr, ansl[curr] - l[curr] + 1, ansl[curr] + ml[curr]);
      } else {
        search2(to, curr, ansr[curr] - mr[curr] + lsum,
                ansr[curr] + r[curr] - rsum - 1);
        lsum += l[to] + ml[to];
        rsum += r[to] + mr[to];
      }
    }
  }
  return;
}
int main() {
  cin.tie(nullptr);
  ios::sync_with_stdio(false);
  int n;
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    cin >> a >> b;
    a--;
    b--;
    v[a].emplace_back(b);
    v[b].emplace_back(a);
  }
  search(0, -1);
  search2(0, -1, 1, 2 * n);
  for (int i = 0; i < n; i++) {
    cout << ansl[i] << " " << ansr[i] << "\n";
  }
  return 0;
}