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 int MOD = 1000000007;
const long long BIG = 1446803456761533460LL;
const int Big = 336860180;
stringstream sss;
const long long int maxn = 500010;
long long int n;
vector<long long int> g[maxn], subt[maxn];
long long int sz[maxn];
pair<long long int, long long int> es[maxn];
void dfs_sz(long long int x, long long int par) {
  sz[x] = 1;
  for (auto y : g[x]) {
    if (y == par) continue;
    subt[x].push_back(y);
    dfs_sz(y, x);
    sz[x] += sz[y];
  }
}
long long int dfs(long long int x, long long int par, long long int l,
                  long long int P) {
  if (P == 0) {
    long long int nn = subt[x].size();
    long long int m = l + nn + 1;
    for (long long int yi = (0); yi < (nn); ++yi) {
      long long int y = subt[x][yi];
      long long int lst = dfs(y, x, m, P ^ 1);
      es[y] = {l + nn - 1 - yi, lst};
      m += sz[y] * 2 - 1;
    }
    return l + nn;
  } else {
    long long int nn = subt[x].size();
    long long int m = l;
    for (long long int yi = (0); yi < (nn); ++yi) {
      long long int y = subt[x][yi];
      long long int lst = dfs(y, x, m, P ^ 1);
      es[y] = {lst, l + sz[x] * 2 - yi - 2};
      m += sz[y] * 2 - 1;
    }
    return l + sz[x] * 2 - nn - 2;
  }
}
void MAIN() {
  cin >> n;
  for (long long int i = (0); i < (n - 1); ++i) {
    long long int a, b;
    cin >> a >> b;
    --a, --b;
    g[a].push_back(b), g[b].push_back(a);
  }
  dfs_sz(0, 0);
  dfs(0, 0, 0, 0);
  es[0] = {subt[0].size(), n * 2 - 1};
  for (long long int i = (0); i < (n); ++i) {
    cout << es[i].first + 1 << ' ' << es[i].second + 1 << '\n';
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  sss << R"(
6
1 2
1 3
3 4
3 5
2 6
    )";
  MAIN();
  return 0;
}