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 long long N = 500500;
vector<long long> g[N];
long long n;
long long r[2 * N], l[2 * N];
long long ans[N][2];
void insToLeftOf(long long x, long long y) {
  long long z = l[y];
  l[x] = z;
  r[x] = y;
  l[y] = x;
  r[z] = x;
}
void dfs(long long v, long long par) {
  if (par != -1) {
    insToLeftOf(2 * v, 2 * par);
    insToLeftOf(2 * v + 1, r[2 * par]);
  }
  for (long long u : g[v]) {
    if (u == par) continue;
    dfs(u, v);
  }
}
int main() {
  cin >> n;
  for (int i = 1; i < n; i++) {
    long long a, b;
    scanf("%lld %lld", &a, &b);
    --a, --b;
    g[a].push_back(b);
    g[b].push_back(a);
  }
  l[0] = 2 * n;
  r[0] = 1;
  l[1] = 0;
  r[1] = 2 * n;
  r[2 * n] = 0;
  l[2 * n] = 1;
  dfs(0, -1);
  long long v = 2 * n;
  for (int i = 1; i <= 2 * n; i++) {
    v = r[v];
    ans[v / 2][v & 1] = i;
  }
  for (int i = 0; i < n; i++) cout << ans[i][0] << " " << ans[i][1] << endl;
  cerr << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << " s.\n";
  return 0;
}