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;
int gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); }
int ppow(int a, int b, int mod) {
  a %= mod;
  int ans = 1 % mod;
  while (b) {
    if (b & 1) ans = (long long)ans * a % mod;
    a = (long long)a * a % mod;
    b >>= 1;
  }
  return ans;
}
int readdd() {
  int x = 0, f = 1;
  char c = getchar();
  while (!isdigit(c) && c != '-') c = getchar();
  if (c == '-') f = -1, c = getchar();
  while (isdigit(c)) x = x * 10 + c - '0', c = getchar();
  return f * x;
}
void printtt(int x) {
  if (x < 0) putchar('-'), x = -x;
  if (x >= 10) printtt(x / 10);
  putchar(x % 10 + '0');
}
bool addd(int a, int b) { return a > b; };
const int maxm = 5e5 + 5;
int l[maxm], r[maxm], tot;
vector<int> g[maxm];
int n;
void dfs(int x, int fa) {
  int len = g[x].size();
  for (int i = len - 1; i >= 0; i--) {
    int v = g[x][i];
    if (v == fa) continue;
    l[v] = ++tot;
  }
  r[x] = ++tot;
  for (int i = 0; i < len; i++) {
    int v = g[x][i];
    if (v == fa) continue;
    dfs(v, x);
  }
}
signed main() {
  scanf("%d", &n);
  for (int i = 1; i < n; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    g[a].push_back(b);
    g[b].push_back(a);
  }
  l[1] = ++tot;
  dfs(1, -1);
  for (int i = 1; i <= n; i++) {
    printf("%d %d\n", l[i], r[i]);
  }
  return 0;
}