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;
vector<int> g[MAXN + 5];
int where = 0;
int L[MAXN + 5];
int R[MAXN + 5];
void dfs(int x, int p) {
  int a = where;
  where += g[x].size() + (p == -1 ? 1 : 0);
  R[x] = where - 1;
  int j = g[x].size() + (p == -1 ? 0 : -1);
  for (int i = 0; i < g[x].size(); ++i) {
    int v = g[x][i];
    if (v != p) {
      --j;
      L[v] = a + j;
      dfs(v, x);
    }
  }
}
void solve() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i < n; ++i) {
    int x, y;
    scanf("%d%d", &x, &y);
    --x, --y;
    g[x].push_back(y);
    g[y].push_back(x);
  }
  L[0] = 0;
  where = 1;
  dfs(0, -1);
  for (int i = 0; i < n; ++i) {
    printf("%d %d\n", L[i] + 1, R[i] + 1);
  }
}
int main() {
  solve();
  return 0;
}