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("Ofast", "unroll-loops")
using namespace std;
int n, t, in[500100], ou[500100], fa[500100];
vector<int> v[500100];
stack<int> st;
void dfs(int a, int f) {
  fa[a] = f;
  for (int i : v[a]) {
    if (i != f) {
      dfs(i, a);
    }
  }
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0);
  cin >> n;
  for (int i = 0, x, y; i + 1 < n; i++) {
    cin >> x >> y, v[x].push_back(y), v[y].push_back(x);
  }
  in[1] = ++t;
  dfs(1, -1);
  st.push(1);
  while (!st.empty()) {
    int now = st.top();
    st.pop();
    for (int i : v[now]) {
      if (i != fa[now]) {
        in[i] = ++t;
        st.push(i);
      }
    }
    ou[now] = ++t;
  }
  for (int i = 1; i <= n; i++) {
    cout << in[i] << ' ' << ou[i] << '\n';
  }
}