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 N = 1e6 + 10;
const int mod = 1e9 + 7;
int power(int a, int b) {
  int res = 1;
  while (b > 0) {
    if (b % 2) {
      res = 1LL * res * a % mod;
    }
    a = 1LL * a * a % mod;
    b >>= 1;
  }
  return res;
}
vector<int> adj[N];
int a[N], b[N];
int dfs(int node, int par, int l, int rr) {
  int cnt = 0;
  for (int child : adj[node]) {
    if (child == par) continue;
    cnt++;
  }
  int r = rr + cnt + 1;
  a[node] = l, b[node] = r;
  cnt = 0;
  for (int child : adj[node]) {
    if (child == par) continue;
    r = max(r, dfs(child, node, b[node] - (++cnt), r));
  }
  return r;
}
int main() {
  int test = 1;
  while (test--) {
    int n, u, v;
    scanf("%d", &n);
    for (int i = 1; i < n; i++) {
      scanf("%d %d", &u, &v);
      adj[u].push_back(v);
      adj[v].push_back(u);
    }
    dfs(1, 0, 1, 1);
    for (int i = 1; i <= n; i++) {
      printf("%d %d\n", a[i], b[i]);
    }
  }
  return 0;
}