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 long long MOD = 1000000007;
const long long MAX = 100005;
const long double PI = 3.14159265359;
const long double G = 9.807;
const long long INF = 1e18;
const long double EPS = 1e-6;
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
bool isprime(long long a) {
  if (a == 2) return 1;
  if (!(a & 1)) return 0;
  for (long long i = 3; i * i <= a; i += 2)
    if (a % i == 0) return 0;
  return 1;
}
long long mpow(long long base, long long exponent, long long modulus) {
  if (modulus == 1) return 0;
  long long result = 1;
  base = base % modulus;
  while (exponent) {
    if (exponent % 2 == 1) result = (result * base) % modulus;
    exponent = exponent >> 1;
    base = (base * base) % modulus;
  }
  return result;
}
long long minv(long long a, long long mod) {
  long long _gcd = gcd(a, mod);
  assert(_gcd == 1);
  return mpow(a, mod - 2, mod);
}
long long n;
vector<long long> g[5 * MAX];
long long L[5 * MAX], R[5 * MAX];
long long time_stamp = 2;
void dfs(long long u, long long p) {
  for (auto v : g[u]) {
    if (v != p) {
      L[v] = time_stamp++;
    }
  }
  R[u] = time_stamp++;
  reverse(g[u].begin(), g[u].end());
  for (auto v : g[u]) {
    if (v != p) {
      dfs(v, u);
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n - 1; ++i) {
    long long u, v;
    cin >> u >> v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  L[1] = 1;
  dfs(1, 0);
  for (int i = 1; i <= n; ++i) {
    cout << L[i] << ' ' << R[i] << '\n';
  }
  return 0;
}