Problem: You are given an unweighted tree with n vertices. Then n - 1 following operations are applied to the tree. A single operation consists of the following steps: 

  1. choose two leaves; 
  2. add the length of the simple path between them to the answer; 
  3. remove one of the chosen leaves from the tree. 



Initial answer (before applying operations) is 0. Obviously after n - 1 such operations the tree will consist of a single vertex. 

Calculate the maximal possible answer you can achieve, and construct a sequence of operations that allows you to achieve this answer!

Input

The first line contains one integer number n (2 ≤ n ≤ 2·105) — the number of vertices in the tree. 

Next n - 1 lines describe the edges of the tree in form ai, bi (1 ≤ ai, bi ≤ n, ai ≠ bi). It is guaranteed that given graph is a tree.

Output

In the first line print one integer number — maximal possible answer. 

In the next n - 1 lines print the operations in order of their applying in format ai, bi, ci, where ai, bi — pair of the leaves that are chosen in the current operation (1 ≤ ai, bi ≤ n), ci (1 ≤ ci ≤ n, ci = ai or ci = bi) — choosen leaf that is removed from the tree in the current operation. 

See the examples for better understanding.

Examples

Input

3
1 2
1 3


Output

3
2 3 3
2 1 1


Input

5
1 2
1 3
2 4
2 5


Output

9
3 5 5
4 3 3
4 1 1
4 2 2

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int N = 200010;
int n, e, S, T, mx;
int hed[N], ver[N << 1], nxt[N << 1];
int d1[N], d2[N];
void add(int x, int y) {
  ver[++e] = y;
  nxt[e] = hed[x];
  hed[x] = e;
}
void dfs1(int x, int pre) {
  d1[x] = d1[pre] + 1;
  if (d1[x] > mx) mx = d1[x], S = x;
  for (int i = hed[x]; i; i = nxt[i]) {
    int y = ver[i];
    if (y == pre) continue;
    dfs1(y, x);
  }
}
void dfs2(int x, int pre) {
  d1[x] = d1[pre] + 1;
  if (d1[x] > mx) mx = d1[x], T = x;
  for (int i = hed[x]; i; i = nxt[i]) {
    int y = ver[i];
    if (y == pre) continue;
    dfs2(y, x);
  }
}
void dfs3(int x, int pre) {
  d2[x] = d2[pre] + 1;
  for (int i = hed[x]; i; i = nxt[i]) {
    int y = ver[i];
    if (y == pre) continue;
    dfs3(y, x);
  }
}
void dfs4(int x, int pre) {
  for (int i = hed[x]; i; i = nxt[i]) {
    int y = ver[i];
    if (y == pre) continue;
    if (d1[y] + d2[y] == d1[T]) continue;
    dfs4(y, x);
  }
  for (int i = hed[x]; i; i = nxt[i]) {
    int y = ver[i];
    if (y == pre) continue;
    if (d1[y] + d2[y] == d1[T]) dfs4(y, x);
  }
  if (d1[x] + d2[x] > d1[T]) {
    if (d1[x] > d2[x])
      printf("%d %d %d\n", S, x, x);
    else
      printf("%d %d %d\n", T, x, x);
  } else {
    if (x != S) printf("%d %d %d\n", S, x, x);
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i < n; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    add(x, y);
    add(y, x);
  }
  dfs1(1, 0);
  mx = 0;
  dfs2(S, 0);
  dfs3(T, 0);
  long long ans = 0;
  for (int i = 1; i <= n; i++) --d1[i], --d2[i];
  for (int i = 1; i <= n; i++) {
    if (d1[i] + d2[i] > d1[T])
      ans += max(d1[i], d2[i]);
    else
      ans += d1[i];
  }
  cout << ans << endl;
  dfs4(S, 0);
  return 0;
}