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 inf = 1e9;
const int mod = 1e9 + 7;
const int maxn = 1e6 + 10;
int n;
vector<int> g[maxn];
int l[maxn], r[maxn];
int pre[maxn], suf[maxn], a[maxn], b[maxn];
void dfs(int u, int fa) {
  int p = 0, st = 0, ed = 0, cnt = 0;
  for (int x : g[u]) {
    if (x == fa) continue;
    cnt++;
    dfs(x, u);
  }
  if (cnt == 0) return;
  for (int x : g[u]) {
    if (x == fa) continue;
    if (!p)
      p = st = ed = x;
    else {
      int tmp = suf[p];
      pre[x] = p;
      suf[p] = x;
      pre[tmp] = b[x];
      suf[b[x]] = tmp;
      p = ed = x;
    }
  }
  suf[u] = st;
  pre[st] = u;
  int tmp = suf[ed];
  suf[ed] = u + n;
  pre[u + n] = ed;
  suf[u + n] = tmp;
  pre[tmp] = u + n;
  b[u] = b[st];
}
int main() {
  scanf("%d", &n);
  for (int i = 1, x, y; i < n; i++) {
    scanf("%d%d", &x, &y);
    g[x].push_back(y);
    g[y].push_back(x);
  }
  for (int i = 1; i <= n; i++) {
    suf[i] = i + n;
    a[i] = i;
    b[i] = i + n;
  }
  for (int i = n + 1; i <= 2 * n; i++) pre[i] = i - n;
  dfs(1, 0);
  int x = 1;
  for (int i = 1; i <= 2 * n; i++) {
    if (x <= n)
      l[x] = i;
    else
      r[x - n] = i;
    x = suf[x];
  }
  for (int i = 1; i <= n; i++) {
    printf("%d %d\n", l[i], r[i]);
  }
  return 0;
}