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;
vector<int> adj[200001];
bool visited[200001];
int distu[200001];
int distv[200001];
int parent[200001];
int initbfs(int i) {
  queue<int> q;
  q.push(i);
  int prev = -1;
  while (!q.empty()) {
    int temp = q.front();
    prev = temp;
    visited[temp] = 1;
    q.pop();
    for (auto j : adj[temp]) {
      if (!visited[j]) {
        visited[j] = 1;
        q.push(j);
      }
    }
  }
  return prev;
}
int initparentbfs(int i) {
  queue<int> q;
  q.push(i);
  int prev = -1;
  while (!q.empty()) {
    int temp = q.front();
    prev = temp;
    visited[temp] = 1;
    q.pop();
    for (auto j : adj[temp]) {
      if (!visited[j]) {
        parent[j] = temp;
        visited[j] = 1;
        q.push(j);
      }
    }
  }
  return prev;
}
void udistcalc(int i) {
  queue<int> q;
  q.push(i);
  distu[i] = 0;
  while (!q.empty()) {
    int temp = q.front();
    visited[temp] = 1;
    q.pop();
    for (auto j : adj[temp]) {
      if (!visited[j]) {
        visited[j] = 1;
        distu[j] = distu[temp] + 1;
        q.push(j);
      }
    }
  }
}
vector<int> dd;
void answercalc(int i) {
  queue<int> q;
  q.push(i);
  while (!q.empty()) {
    int temp = q.front();
    visited[temp] = 1;
    dd.push_back(temp);
    q.pop();
    for (auto j : adj[temp]) {
      if (!visited[j]) {
        visited[j] = 1;
        q.push(j);
      }
    }
  }
}
void vdistcalc(int i) {
  queue<int> q;
  q.push(i);
  distv[i] = 0;
  while (!q.empty()) {
    int temp = q.front();
    visited[temp] = 1;
    q.pop();
    for (auto j : adj[temp]) {
      if (!visited[j]) {
        visited[j] = 1;
        distv[j] = distv[temp] + 1;
        q.push(j);
      }
    }
  }
}
int arr[200001];
int main(int argc, char const *argv[]) {
  ios_base::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
  int n;
  cin >> n;
  int x, y;
  for (int i = 1; i <= n - 1; i++) {
    cin >> x >> y;
    adj[x].push_back(y);
    adj[y].push_back(x);
  }
  for (int i = 1; i <= n; i++) visited[i] = 0;
  int u = initbfs(1);
  for (int i = 1; i <= n; i++) visited[i] = 0;
  int v = initparentbfs(u);
  for (int i = 1; i <= n; i++) visited[i] = 0;
  udistcalc(u);
  for (int i = 1; i <= n; i++) visited[i] = 0;
  vdistcalc(v);
  long long ans = 0;
  vector<int> s;
  int i = v;
  while (i != u) {
    s.push_back(i);
    i = parent[i];
  }
  s.push_back(u);
  vector<int>::iterator it;
  vector<int>::iterator pos;
  vector<pair<int, pair<int, int>>> final;
  for (it = s.begin(); it != s.end(); it++) {
    arr[*it] = 1;
    if (*it == u) pos = it;
  }
  s.erase(pos);
  for (int i = 1; i <= n; i++) visited[i] = 0;
  answercalc(u);
  reverse(dd.begin(), dd.end());
  for (int i = 0; i < dd.size(); i++) {
    if (!arr[dd[i]]) {
      if (distu[dd[i]] >= distv[dd[i]]) {
        ans += distu[dd[i]];
        final.push_back({dd[i], {u, dd[i]}});
      } else {
        ans += distv[dd[i]];
        final.push_back({dd[i], {v, dd[i]}});
      }
    }
  }
  for (it = s.begin(); it != s.end(); it++) final.push_back({*it, {u, *it}});
  ans += (1ll * distu[v] * (distu[v] + 1) / 2);
  cout << ans << endl;
  for (int i = 0; i < final.size(); i++) {
    cout << final[i].first << " " << final[i].second.first << " "
         << final[i].second.second << endl;
  }
  return 0;
}