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 limite = 1000000;
void maximiza(pair<int, int> &a, pair<int, int> b) { a = max(a, b); }
int lenmax, umax, vmax;
int n;
vector<int> g[limite];
pair<int, int> altura[limite];
vector<pair<int, int> > alturas[limite];
void calcula(int u, int p) {
  altura[u] = pair<int, int>(0, u);
  vector<pair<int, int> > &alu = alturas[u];
  alu.push_back(altura[u]);
  vector<int> &ar = g[u];
  for (int i = 0; i < int(ar.size()); i++) {
    int v = ar[i];
    if (v != p) {
      calcula(v, u);
      maximiza(altura[u],
               pair<int, int>(1 + altura[v].first, altura[v].second));
      alu.push_back(pair<int, int>(1 + altura[v].first, altura[v].second));
    }
  }
  sort(alu.begin(), alu.end());
  if (int(alu.size()) > 1) {
    int i1 = int(alu.size()) - 2;
    int i2 = i1 + 1;
    if (alu[i1].first + alu[i2].first > lenmax) {
      lenmax = alu[i1].first + alu[i2].first;
      umax = alu[i1].second;
      vmax = alu[i2].second;
    }
  }
}
int padre[limite];
void eliminapadre(int u, int p) {
  padre[u] = p;
  vector<int> &ar = g[u];
  vector<int> nextar;
  for (int i = 0; i < int(ar.size()); i++) {
    int v = ar[i];
    if (v != p) {
      eliminapadre(v, u);
      nextar.push_back(v);
    }
  }
  ar = nextar;
}
long long int acum = 0;
vector<pair<int, int> > sol;
void elimina(int u, int len, int uref) {
  vector<int> &ar = g[u];
  for (int i = 0; i < int(ar.size()); i++) {
    int v = ar[i];
    elimina(v, len + 1, uref);
  }
  acum += len;
  sol.push_back(pair<int, int>(uref, u));
}
vector<int> lista;
void eliminasides(int u, int uant, int dist) {
  if (u == 0) return;
  lista.push_back(u);
  vector<int> &ar = g[u];
  for (int i = 0; i < int(ar.size()); i++) {
    int v = ar[i];
    if (v != uant) {
      if (dist > lenmax - dist)
        elimina(v, 1 + dist, vmax);
      else
        elimina(v, 1 + lenmax - dist, umax);
    }
  }
  eliminasides(padre[u], u, dist + 1);
}
int main() {
  ios::sync_with_stdio(false);
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    cin >> u >> v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  calcula(1, 0);
  eliminapadre(umax, 0);
  eliminasides(vmax, 0, 0);
  for (int i = 0; i < int(lista.size()) - 1; i++) {
    acum += int(lista.size()) - 1 - i;
    sol.push_back(pair<int, int>(lista.back(), lista[i]));
  }
  cout << acum << endl;
  for (int i = 0; i < int(sol.size()); i++)
    cout << sol[i].first << " " << sol[i].second << " " << sol[i].second
         << endl;
}