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 int maxN = 5e5 + 7;
int N, head[maxN], cnt;
struct Eddge {
  int nex, to;
  Eddge(int a = -1, int b = 0) : nex(a), to(b) {}
} edge[maxN << 1];
inline void addEddge(int u, int v) {
  edge[cnt] = Eddge(head[u], v);
  head[u] = cnt++;
}
inline void _add(int u, int v) {
  addEddge(u, v);
  addEddge(v, u);
}
int siz[maxN], fa[maxN], dir[maxN];
void pre_dfs(int u, int father) {
  siz[u] = 1;
  fa[u] = father;
  dir[u] = 0;
  for (int i = head[u], v; ~i; i = edge[i].nex) {
    v = edge[i].to;
    if (v == father) continue;
    pre_dfs(v, u);
    dir[u]++;
    siz[u] += siz[v];
  }
}
struct node {
  int l, r;
  node(int a = 0, int b = 0) : l(a), r(b) {}
  inline void Out() { printf("%d %d\n", l, r); }
} ans[maxN];
int L;
void dfs(int u, int fail) {
  ans[u] = node(L - dir[u], fail);
  L = L - 1 - dir[u];
  int Beg = L + 2;
  for (int i = head[u], v; ~i; i = edge[i].nex) {
    v = edge[i].to;
    if (v == fa[u]) continue;
    dfs(v, Beg++);
  }
}
inline void init() {
  cnt = 0;
  for (int i = 1; i <= N; i++) head[i] = -1;
}
int main() {
  scanf("%d", &N);
  init();
  for (int i = 1, u, v; i < N; i++) {
    scanf("%d%d", &u, &v);
    _add(u, v);
  }
  pre_dfs(1, 0);
  L = (N << 1) - 1;
  dfs(1, N << 1);
  for (int i = 1; i <= N; i++) ans[i].Out();
  return 0;
}