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;
mt19937 rnd(chrono::steady_clock::now().time_since_epoch().count());
vector<vector<int>> g;
vector<bool> used;
vector<int> pr;
int maxd = 0;
int d0 = 0;
int d1 = 0;
void dfs_s(int v, int d) {
  used[v] = true;
  if (d > maxd) {
    maxd = d;
    d0 = v;
  }
  for (auto u : g[v]) {
    if (!used[u]) {
      pr[u] = v;
      dfs_s(u, d + 1);
    }
  }
}
unsigned long long dfs_cnter(int v, int d, int dplus) {
  used[v] = true;
  unsigned long long ans = 0;
  for (auto u : g[v]) {
    if (!used[u]) ans += dfs_cnter(u, d + 1, dplus);
  }
  ans += d + dplus;
  return ans;
}
void dfs_print(int v, int par) {
  used[v] = true;
  for (auto u : g[v]) {
    if (!used[u]) dfs_print(u, par);
  }
  cout << v + 1 << " " << par + 1 << " " << v + 1 << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  g.resize(n, vector<int>(0));
  used.resize(n, false);
  pr.resize(n, 0);
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    cin >> u >> v;
    g[u - 1].push_back(v - 1);
    g[v - 1].push_back(u - 1);
  }
  dfs_s(0, 0);
  maxd = 0;
  pr[d0] = -1;
  int da = d0;
  fill(used.begin(), used.end(), false);
  dfs_s(d0, 0);
  int db = d0;
  vector<int> diag(0);
  int curr = db;
  while (curr != -1) {
    diag.push_back(curr);
    curr = pr[curr];
  }
  reverse(diag.begin(), diag.end());
  fill(used.begin(), used.end(), false);
  for (auto i : diag) used[i] = true;
  unsigned long long ans = 0;
  for (int i = 0; i < diag.size(); i++) {
    int dcnt = i;
    int farest = da;
    if (diag.size() - 1 - i > i) {
      dcnt = diag.size() - 1 - i;
      farest = db;
    }
    int cur = diag[i];
    for (auto u : g[cur])
      if (!used[u]) ans += dfs_cnter(u, 1, dcnt);
  }
  unsigned long long d = diag.size();
  ans += d * (d - 1) / 2;
  cout << ans << endl;
  fill(used.begin(), used.end(), false);
  for (auto i : diag) used[i] = true;
  for (int i = 0; i < diag.size(); i++) {
    int farest = da;
    if (diag.size() - 1 - i > i) {
      farest = db;
    }
    int cur = diag[i];
    for (auto u : g[cur])
      if (!used[u]) dfs_print(u, farest);
  }
  for (int i = 0; i < d - 1; i++)
    cout << diag[i] + 1 << " " << db + 1 << " " << diag[i] + 1 << endl;
  return 0;
}