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;
long long power(long long x, long long y, long long z) {
  long long ret = 1;
  while (y > 0) {
    if (y & 1) ret = (ret * x) % z;
    x = (x * x) % z;
    y >>= 1;
  }
  return ret;
}
const long long N = 5e5 + 5;
long long L[N], R[N], cnt = 1;
vector<long long> V[N];
void dfs(long long x, long long p) {
  for (auto u : V[x]) {
    if (u == p) continue;
    L[u] = cnt++;
  }
  R[x] = cnt++;
  reverse((V[x]).begin(), (V[x]).end());
  for (auto u : V[x]) {
    if (u == p) continue;
    dfs(u, x);
  }
}
int main() {
  long long n;
  scanf("%lld", &n);
  for (long long i = 1; i <= n - 1; ++i) {
    long long x, y;
    scanf("%lld%lld", &x, &y);
    V[x].push_back(y);
    V[y].push_back(x);
  }
  L[1] = cnt++;
  dfs(1, 0);
  for (long long i = 1; i <= n; ++i) printf("%lld %lld\n", L[i], R[i]);
  return 0;
}