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 = 500100;
long long read() {
  long long x = 0, f = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = x * 10 + c - '0';
    c = getchar();
  }
  return x * f;
}
vector<int> e[N];
int l[N], r[N], now;
void dfs(int u, int fa) {
  for (vector<int>::iterator it = e[u].begin(); it != e[u].end(); ++it) {
    if (fa == *it) continue;
    l[*it] = ++now;
  }
  r[u] = ++now;
  int k = e[u].size();
  reverse(e[u].begin(), e[u].end());
  for (vector<int>::iterator it = e[u].begin(); it != e[u].end(); ++it) {
    if (fa == *it) continue;
    dfs(*it, u);
  }
}
int main() {
  int n = read();
  for (int i = 1; i < n; ++i) {
    int u = read(), v = read();
    e[u].push_back(v);
    e[v].push_back(u);
  }
  l[1] = ++now;
  dfs(1, 0);
  for (int i = 1; i <= n; ++i) printf("%d %d\n", l[i], r[i]);
  return 0;
}