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>
#pragma GCC optimize("O3")
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
using namespace std;
long long N = (long long)1e9 + 7, M = 20;
long long i, j, x, y, z, m, n, k, l, r, f, mi, ans, x2, y2, ma, t, a[1110000],
    b[1110000], d[410000], p[1110000];
string s, s2;
vector<long long> v[1110000], v2, v3;
set<long long> st;
void dfs(int x, int y) {
  int l = 0, r = 0;
  for (int i = 0; i < v[x].size(); i++) {
    int to = v[x][i];
    if (to == y) continue;
    l++;
  }
  for (int i = 0; i < v[x].size(); i++) {
    int to = v[x][i];
    if (to == y) continue;
    r++;
    a[to] = k + l - r + 1;
  }
  k += l;
  k++;
  b[x] = k;
  for (int i = 0; i < v[x].size(); i++) {
    int to = v[x][i];
    if (to == y) continue;
    dfs(to, x);
  }
}
int main() {
  scanf("%d", &n);
  for (i = 1; i < n; i++) {
    scanf("%d%d", &x, &y);
    v[x].push_back(y);
    v[y].push_back(x);
  }
  k = 1;
  a[1] = 1;
  dfs(1, 0);
  for (i = 1; i <= n; i++) {
    printf("%d ", a[i]);
    printf("%d\n", b[i]);
  }
}