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 n, m, a[1000100], cnt, cnt2, flag[1000100], cnt3, tot, l[1000100],
    r[1000100];
vector<int> q[1000100];
struct edge {
  int l, r;
} d[1000100];
struct node {
  int v;
  node *next;
} pool[2 * 1000100], *h[1000100];
void addedge(int u, int v) {
  node *p = &pool[++cnt], *q = &pool[++cnt];
  p->v = v;
  p->next = h[u];
  h[u] = p;
  q->v = u;
  q->next = h[v];
  h[v] = q;
}
void dfs(int u, int fa) {
  for (node *p = h[u]; p; p = p->next) {
    if (p->v == fa) continue;
    q[u].push_back(p->v);
    l[p->v] = ++tot;
  }
  r[u] = ++tot;
  for (int i = q[u].size() - 1; i >= 0; i--) {
    dfs(q[u][i], u);
  }
}
int main() {
  scanf("%d", &n);
  int u, v;
  for (int i = 1; i <= n - 1; i++) {
    scanf("%d%d", &u, &v);
    addedge(u, v);
  }
  l[1] = 1;
  tot = 1;
  dfs(1, 0);
  for (int i = 1; i <= n; i++) {
    printf("%d %d\n", l[i], r[i]);
  }
  return 0;
}