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>
inline long long ceil_div(long long a, long long b) { return (a + b - 1) / b; }
using namespace std;
inline int qr() {
  int f = 0, fu = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') fu = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    f = (f << 3) + (f << 1) + (c & 15);
    c = getchar();
  }
  return f * fu;
}
const int N = 5e5 + 10;
int head[N], ver[N << 1], Next[N << 1], tot;
int siz[N], son[N], n;
pair<int, int> ans[N];
inline void add(int x, int y) {
  ver[++tot] = y;
  Next[tot] = head[x];
  head[x] = tot;
}
void dfs(int x, int f) {
  siz[x] = 1;
  for (int i = head[x], y; i; i = Next[i]) {
    if ((y = ver[i]) == f) continue;
    dfs(y, x), siz[x] += siz[y];
  }
}
void solve(int x, int f, int l, int r) {
  ans[x] = {l, r};
  for (int i = head[x], nr = l + 1, nl = l, y; i; i = Next[i]) {
    if ((y = ver[i]) == f) continue;
    solve(y, x, nl - son[y] - 1, nr), nl -= siz[y] * 2 - 1, nr++;
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 1, x, y; i <= n - 1; i++) {
    scanf("%d%d", &x, &y);
    add(x, y), add(y, x);
    son[x]++, son[y]++;
  }
  for (int i = 2; i <= n; i++) son[i]--;
  dfs(1, 0);
  solve(1, 0, 2 * n - son[1] - 1, 2 * n);
  for (int i = 1; i <= n; i++) printf("%d %d\n", ans[i].first, ans[i].second);
  return 0;
}