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 = 567890;
vector<int> G[MAXN];
int sz[MAXN];
pair<int, int> ans[MAXN];
int dfs(int cur, int p) {
  int cur_sz = 0;
  for (int i : G[cur]) {
    if (i == p) continue;
    cur_sz += dfs(i, cur);
  }
  sz[cur] = cur_sz;
  return cur_sz + 1;
}
void solve(int cur, int par, int ed) {
  int idx = ed - 1, offs = 0;
  for (int i = 0; i < (int)G[cur].size(); i++) {
    int to = G[cur][i];
    if (to == par) continue;
    ans[to] = make_pair(idx, idx + offs + (int)G[to].size() - 1 + 2);
    solve(to, cur, idx + offs + (int)G[to].size() - 1 + 2);
    offs += ((sz[to] + 1) * 2);
    idx--;
  }
}
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);
  }
  dfs(1, 1);
  ans[1] = make_pair(1, (int)G[1].size() + 2);
  solve(1, 1, (int)G[1].size() + 2);
  for (int i = 1; i <= N; i++) printf("%d %d\n", ans[i].first, ans[i].second);
  return 0;
}