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 maxn = 5e5 + 10;
vector<int> g[maxn];
int n, ind, tmp[maxn], l[maxn], r[maxn];
inline void dfs(int u, int par) {
  for (int i = g[u].size() - 1; ~i; i--) {
    int mid = g[u][i];
    if (mid != par) l[mid] = ++ind;
  }
  r[u] = ++ind;
  for (int mid : g[u])
    if (mid != par) dfs(mid, u);
}
int main() {
  ios::sync_with_stdio(0);
  cin >> n;
  for (int i = 1, u, v; i < n; i++)
    cin >> u >> v, g[u].push_back(v), g[v].push_back(u);
  l[1] = ++ind, dfs(1, 1);
  for (int i = 1; i <= n; i++) cout << l[i] << " " << r[i] << endl;
  return 0;
}