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>
int faster_in() {
  int r = 0, c;
  for (c = getchar(); c <= 32; c = getchar())
    ;
  if (c == '-') return -faster_in();
  for (; c > 32; r = (r << 1) + (r << 3) + c - '0', c = getchar())
    ;
  return r;
}
using namespace std;
const int INF = int(1e9 + 7);
const int tam = 500010;
vector<int> G[tam];
int L[tam], R[tam];
int n;
int minL[tam];
void dfs(int v, int p, int minVal) {
  int children = G[v].size() - (p != -1);
  L[v] = minVal - children;
  minL[v] = L[v];
  int cnt = 0;
  for (int u : G[v]) {
    if (u == p) continue;
    R[u] = L[v] + (++cnt);
    dfs(u, v, minL[v] - 1);
    minL[v] = min(minL[v], minL[u]);
  }
}
int main() {
  std::ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  for (int i = 0; i < n - 1; ++i) {
    int a, b;
    cin >> a >> b;
    a--;
    b--;
    G[a].emplace_back(b);
    G[b].emplace_back(a);
  }
  R[0] = 2 * n;
  dfs(0, -1, 2 * n - 1);
  for (int i = 0; i < n; ++i) {
    cout << L[i] << ' ' << R[i] << '\n';
  }
  return 0;
}