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 INFL = 1LL << 60;
const int INF = 1 << 30;
long long MOD = 1000000007;
vector<long long> dy = {1, 0, 1};
vector<long long> dx = {0, 1, 1};
void pres(double A, long long x = 10) {
  cout << fixed << setprecision(x) << A << endl;
}
void BinarySay(long long x, long long y = 60) {
  for (long long i = 0; i < (y); i++) cout << (x >> (y - 1 - i) & 1);
  cout << endl;
}
long long cnt_bit(long long x) { return __builtin_popcountll(x); }
long long pow_long(long long x, long long k) {
  long long res = 1;
  while (k > 0) {
    if (k % 2) res *= x;
    x *= x;
    k /= 2;
  }
  return res;
}
long long pow_mod(long long x, long long k) {
  long long res = 1;
  while (k > 0) {
    if (k % 2) {
      res *= x;
      res %= MOD;
    }
    x *= x;
    x %= MOD;
    k /= 2;
  }
  return res;
}
long long inverse(long long x) { return pow_mod(x, MOD - 2); };
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long lcm(long long x, long long y) { return x / gcd(x, y) * y; };
long long N;
vector<long long> E[500010];
long long u[500010], v[500010];
long long x = 1;
void dfs(long long now, long long from) {
  for (long long i = 0; i < (E[now].size()); i++) {
    long long to = E[now][i];
    if (to == from) continue;
    u[to] = x;
    x++;
  }
  v[now] = x;
  x++;
  for (long long i = (E[now].size()) - 1; i >= 0; i--) {
    long long to = E[now][i];
    if (to == from) continue;
    dfs(to, now);
  }
}
void solve() {
  cin >> N;
  if (N == 1) {
    cout << 1 << " " << 2 << endl;
    return;
  }
  for (long long i = 0; i < (N - 1); i++) {
    long long a, b;
    cin >> a >> b;
    a--, b--;
    E[a].push_back(b);
    E[b].push_back(a);
  }
  u[0] = x;
  x++;
  dfs(0, -1);
  for (long long i = 0; i < (N); i++) {
    cout << u[i] << " " << v[i] << endl;
  }
}
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  solve();
}