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>
const int MAXN = 5e5 + 5;
struct Edge {
  int next, to;
} e[MAXN << 1];
int head[MAXN], etot = 0;
inline void add(int u, int v) {
  e[++etot] = (Edge){head[u], v};
  head[u] = etot;
}
struct Node {
  int lst, nxt;
} p[MAXN * 2];
inline void putlst(int v, int u) {
  p[v].lst = p[u].lst;
  p[p[u].lst].nxt = v;
  p[v].nxt = u;
  p[u].lst = v;
}
inline void putnxt(int v, int u) {
  p[v].nxt = p[u].nxt;
  p[p[u].nxt].lst = v;
  p[v].lst = u;
  p[u].nxt = v;
}
int n;
void dfs(int u, int fa) {
  for (int i = head[u]; i; i = e[i].next) {
    int v = e[i].to;
    if (v == fa) continue;
    putlst(v, u + n);
    putnxt(v + n, u + n);
    dfs(v, u);
  }
}
int pos[MAXN * 2];
int main(void) {
  scanf("%d", &n);
  for (int i = 1; i < n; ++i) {
    int u, v;
    scanf("%d%d", &u, &v);
    add(u, v);
    add(v, u);
  }
  p[1].nxt = p[0].lst = n + 1;
  p[n + 1].lst = p[0].nxt = 1;
  dfs(1, 0);
  int now = 0;
  for (int i = 1; i; i = p[i].nxt) pos[i] = ++now;
  for (int i = 1; i <= n; ++i) printf("%d %d\n", pos[i], pos[i + n]);
  return 0;
}