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 = 500010;
vector<int> G[maxn];
int n, cnt = 1;
int ans[maxn][2];
vector<pair<int, int> > res;
void dfs2(int cur, int pre, int len) {
  int size;
  if (cur == 1)
    size = G[1].size();
  else
    size = G[cur].size() - 1;
  cnt += size + 1;
  ans[cur][0] = len, ans[cur][1] = cnt;
  int p = cnt;
  for (int i = 0; i < G[cur].size(); i++) {
    if (G[cur][i] != pre) dfs2(G[cur][i], cur, --p);
  }
}
int main() {
  ios::sync_with_stdio(false);
  while (cin >> n) {
    int a, b;
    res.clear();
    for (int i = 1; i < n; i++) {
      cin >> a >> b;
      G[a].push_back(b);
      G[b].push_back(a);
    }
    dfs2(1, 0, 1);
    for (int i = 1; i <= n; i++) cout << ans[i][0] << " " << ans[i][1] << endl;
  }
  return 0;
}