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 target("avx2")
#pragma GCC optimization("unroll-loops")
using namespace std;
using ll = long long;
using P = pair<int, int>;
constexpr ll INF = (1LL << 30) - 1LL;
constexpr ll LINF = (1LL << 60) - 1LL;
constexpr ll MOD = 1e9 + 7;
template <typename T>
void chmin(T &a, T b) {
  a = min(a, b);
}
template <typename T>
void chmax(T &a, T b) {
  a = max(a, b);
}
vector<int> G[500010];
int l[500010], r[500010];
int dfs(const int &i, const int &p, const int &s) {
  if (p == -1) {
    l[i] = 1;
    r[i] = G[i].size() + 2;
  }
  int sum = 0;
  int co = G[i].size();
  for (auto &e : G[i]) {
    if (e == p) {
      continue;
    }
    l[e] = l[i] + s + co;
    r[e] = l[e] + G[e].size() + 1 + sum;
    int ss = dfs(e, i, sum);
    sum += ss;
    --co;
  }
  return sum + 2;
}
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  int n;
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    cin >> u >> v;
    --u;
    --v;
    G[u].emplace_back(v);
    G[v].emplace_back(u);
  }
  dfs(0, -1, 0);
  for (int i = 0; i < n; i++) {
    cout << l[i] << " " << r[i] << "\n";
  }
}