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 INF = 0x3f3f3f3f;
const long long LLINF = 1e18;
const int maxN = 5e5 + 5;
vector<int> g[maxN];
bool vis[maxN];
queue<int> order;
stack<int> open;
pair<int, int> out[maxN];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    scanf("%d %d", &a, &b);
    g[a].push_back(b);
    g[b].push_back(a);
  }
  vis[1] = true;
  open.push(1);
  order.push(1);
  while (!open.empty()) {
    int x = open.top();
    open.pop();
    for (int y : g[x]) {
      if (vis[y]) continue;
      vis[y] = true;
      open.push(y);
      order.push(y);
    }
    order.push(-x);
  }
  int aux;
  int ind = 1;
  while (!order.empty()) {
    aux = order.front();
    if (aux > 0)
      out[aux].first = ind++;
    else
      out[-aux].second = ind++;
    order.pop();
  }
  for (int i = 1; i < n + 1; i++)
    printf("%d %d\n", out[i].first, out[i].second);
  return 0;
}