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 N = 500005;
int n, du[N], l[N], r[N], head[N], cnt, son[N], sz[N];
struct Edge {
  int v, nx;
} e[N << 1];
inline int read() {
  int ret = 0, ff = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') ff = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    ret = ret * 10 + (ch ^ 48);
    ch = getchar();
  }
  return ret * ff;
}
void write(int x) {
  if (x < 0) {
    x = -x, putchar('-');
  }
  if (x > 9) write(x / 10);
  putchar(x % 10 + 48);
}
void writeln(int x) { write(x), puts(""); }
void writesp(int x) { write(x), putchar(' '); }
void add(int x, int y) {
  e[++cnt].v = y;
  e[cnt].nx = head[x];
  head[x] = cnt;
}
void dfs(int now, int fa, int L, int R) {
  l[now] = L, r[now] = R;
  int t = R, szbro = 0;
  for (int i = head[now]; i; i = e[i].nx) {
    int v = e[i].v;
    if (v == fa) continue;
    t--;
    dfs(v, now, t, t + szbro * 2 + du[v] + 1);
    szbro += sz[v];
  }
}
void init(int now, int fa) {
  sz[now] = 1;
  for (int i = head[now]; i; i = e[i].nx) {
    int v = e[i].v;
    if (v == fa) continue;
    init(v, now);
    son[now]++;
    sz[now] += sz[v];
  }
}
signed main() {
  n = read();
  for (int i = 1; i < n; i++) {
    int x = read(), y = read();
    du[x]++, du[y]++;
    add(x, y), add(y, x);
  }
  init(1, 0);
  dfs(1, 0, 1, du[1] + 2);
  for (int i = 1; i <= n; i++) writesp(l[i]), writeln(r[i]);
  return 0;
}