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 + 50;
vector<int> g[maxn];
int vis[maxn];
int cap[maxn];
int ll[maxn], rr[maxn];
int lm;
void dfs1(int now) {
  vis[now] = 1;
  int ans = 0;
  for (int i = 0; i < g[now].size(); i++) {
    if (!vis[g[now][i]]) {
      dfs1(g[now][i]);
      ans++;
    }
  }
  cap[now] = ans;
  vis[now] = 0;
}
void dfs2(int now, int l, int r) {
  lm = min(lm, l);
  ll[now] = l;
  rr[now] = r;
  vis[now] = 1;
  int cnt = 0;
  for (int i = 0; i < g[now].size(); i++) {
    if (!vis[g[now][i]]) {
      dfs2(g[now][i], lm - cap[g[now][i]] - 1, l + (++cnt));
    }
  }
  vis[now] = 0;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n - 1; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    g[a].push_back(b);
    g[b].push_back(a);
  }
  dfs1(1);
  lm = 2 * n - cap[1] - 1;
  dfs2(1, 2 * n - cap[1] - 1, 2 * n);
  for (int i = 1; i <= n; i++) {
    printf("%d %d\n", ll[i], rr[i]);
  }
}