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 mxx = 1e6 + 60;
int l[mxx], r[mxx], chl[mxx], chr[mxx];
int n;
vector<int> v[mxx];
void fil(int nod, int par) {
  l[nod] = 0, r[nod] = 1;
  int curl = -1, curr = +1;
  for (auto x : v[nod]) {
    if (x == par) continue;
    fil(x, nod);
    r[x] = curr;
    r[nod]++;
    curr++;
    l[x] = curl - chr[x];
    curl = l[x] + chl[x] - 1;
  }
  chl[nod] = curl + 1;
  chr[nod] = curr - 1;
}
void build(int nod, int par) {
  for (auto x : v[nod]) {
    if (x == par) continue;
    l[x] += l[nod];
    r[x] += l[nod];
    build(x, nod);
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    cin >> a >> b;
    v[a].push_back(b);
    v[b].push_back(a);
  }
  fil(1, 0);
  build(1, 0);
  int mn = mxx;
  for (int i = 1; i <= n; i++) {
    mn = min(mn, l[i]);
  }
  mn = 1 - mn;
  for (int i = 1; i <= n; i++) {
    cout << l[i] + mn << ' ' << r[i] + mn << endl;
  }
  return 0;
}