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;
struct edge {
  int to, next;
} net[500039 << 1];
int n, tot, head[500039], siz[500039], son[500039], L[500039], R[500039];
void add(int u, int v) {
  net[++tot] = (edge){v, head[u]}, head[u] = tot;
  net[++tot] = (edge){u, head[v]}, head[v] = tot;
}
void dfs(int pos, int pre) {
  siz[pos] = son[pos] = 1;
  for (int a = head[pos]; a; a = net[a].next) {
    if (net[a].to != pre) {
      dfs(net[a].to, pos);
      siz[pos] += siz[net[a].to];
      son[pos]++;
    }
  }
}
void solve(int pos, int pre, int l, int r) {
  L[pos] = l, R[pos] = r, l = r;
  for (int a = head[pos]; a; a = net[a].next) {
    if (net[a].to != pre) {
      solve(net[a].to, pos, l - 1, r + son[net[a].to]);
      l--, r += siz[net[a].to] * 2 - 1;
    }
  }
}
int main() {
  scanf("%d", &n);
  for (int a = 1, b, c; a < n; a++) {
    scanf("%d%d", &b, &c);
    add(b, c);
  }
  dfs(1, 0);
  solve(1, 0, 1, son[1] + 1);
  for (int a = 1; a <= n; a++) {
    printf("%d %d\n", L[a], R[a]);
  }
  return 0;
}