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 = 1e9 + 7, N = 5e5 + 7, M = 2e6 + 7, INF = INT_MAX / 10;
long long powe(long long x, long long y) {
  x = x % mod, y = y % (mod - 1);
  long long ans = 1;
  while (y > 0) {
    if (y & 1) {
      ans = (1ll * x * ans) % mod;
    }
    y >>= 1;
    x = (1ll * x * x) % mod;
  }
  return ans;
}
vector<vector<int>> edg(N);
vector<bool> vis(N);
vector<int> siz(N);
vector<pair<int, int>> ans(N);
int n;
int calsiz(int node) {
  vis[node] = true;
  int ans = 1;
  for (auto& x : edg[node]) {
    if (!vis[x]) {
      ans += calsiz(x);
    }
  }
  siz[node] = ans;
  return ans;
}
void dfs(int parentLoc, int start, int node) {
  vis[node] = true;
  int branches = 0;
  int loc1 = start, loc2 = start + 2 * siz[node] - 1;
  for (auto& x : edg[node]) {
    if (!vis[x]) {
      loc2 -= 2 * siz[x] - 1;
      dfs(loc1, loc2, x);
      loc1++;
      branches++;
    }
  }
  ans[node] = (make_pair(parentLoc, start + branches));
}
void solve() {
  int a, b;
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    cin >> a >> b;
    edg[a].push_back(b);
    edg[b].push_back(a);
  }
  calsiz(1);
  vis.assign(N, 0);
  dfs(1, 2, 1);
  for (int i = 1; i <= n; i++) {
    auto x = ans[i];
    cout << x.first << " " << x.second << "\n";
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  int t = 1;
  while (t--) {
    solve();
  }
  return 0;
}