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 int N = 5e5 + 5, mod = 31, bit = 60;
vector<long long int> adj[N];
long long int s[N], e[N], timer = 0;
void dfs(long long int u, long long int p) {
  for (auto v : adj[u]) {
    if (v != p) {
      s[v] = ++timer;
    }
  }
  reverse(adj[u].begin(), adj[u].end());
  e[u] = ++timer;
  for (auto v : adj[u]) {
    if (v != p) {
      dfs(v, u);
    }
  }
}
signed main() {
  ios_base::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
  long long int pro = 1, temp, t, i, j, l, r, n, m, x, y, z, mid, k, rem,
                carry = 0, ind, ans = 0, mx = -LONG_LONG_MAX,
                mn = LONG_LONG_MAX, cnt = 0, curr = 0, prev, next, sum = 0,
                i1 = -1, i2 = -1;
  bool flag;
  cin >> n;
  for (i = 1; i < n; i++) {
    cin >> x >> y;
    adj[x].push_back(y);
    adj[y].push_back(x);
  }
  s[1] = ++timer;
  dfs(1, 0);
  for (i = 1; i <= n; i++) {
    cout << s[i] << ' ' << e[i] << '\n';
  }
  return 0;
}