#include <bits/stdc++.h>
using namespace std;
template <typename T, typename U>
inline void smin(T &a, U b) {
  if (a > b) a = b;
}
template <typename T, typename U>
inline void smax(T &a, U b) {
  if (a < b) a = b;
}
template <class T>
inline void gn(T &first) {
  char c, sg = 0;
  while (c = getchar(), (c > '9' || c < '0') && c != '-')
    ;
  for ((c == '-' ? sg = 1, c = getchar() : 0), first = 0; c >= '0' && c <= '9';
       c = getchar())
    first = (first << 1) + (first << 3) + c - '0';
  if (sg) first = -first;
}
template <class T, class T1>
inline void gn(T &first, T1 &second) {
  gn(first);
  gn(second);
}
template <class T, class T1, class T2>
inline void gn(T &first, T1 &second, T2 &z) {
  gn(first);
  gn(second);
  gn(z);
}
template <class T>
inline void print(T first) {
  if (first < 0) {
    putchar('-');
    return print(-first);
  }
  if (first < 10) {
    putchar('0' + first);
    return;
  }
  print(first / 10);
  putchar(first % 10 + '0');
}
template <class T>
inline void printsp(T first) {
  print(first);
  putchar(' ');
}
template <class T>
inline void println(T first) {
  print(first);
  putchar('\n');
}
template <class T, class U>
inline void print(T first, U second) {
  printsp(first);
  println(second);
}
template <class T, class U, class V>
inline void print(T first, U second, V z) {
  printsp(first);
  printsp(second);
  println(z);
}
int power(int a, int b, int m, int ans = 1) {
  for (; b; b >>= 1, a = 1LL * a * a % m)
    if (b & 1) ans = 1LL * ans * a % m;
  return ans;
}
vector<int> adj[200200];
pair<int, int> dfs(int u, int d, int pa = -1) {
  pair<int, int> first = pair<int, int>(d, u);
  for (int i = 0; i < adj[u].size(); i++)
    if (adj[u][i] != pa) smax(first, dfs(adj[u][i], d + 1, u));
  return first;
}
long long ans;
vector<pair<int, int> > res;
int vst[200200], pre[200200];
void solve(int u, pair<int, int> d, int pa = -1) {
  for (int i = 0; i < adj[u].size(); i++) {
    int v = adj[u][i];
    if (v == pa || vst[v]) continue;
    res.push_back(pair<int, int>(d.first, v));
    ans += d.second + 1;
    solve(v, pair<int, int>(d.first, d.second + 1), u);
  }
}
int main() {
  int n, u, v;
  gn(n);
  if (n == 2) {
    puts("1");
    puts("1 2 2");
    return 0;
  }
  for (int i = 1; i < n; i++) {
    gn(u, v);
    if (u > v) swap(u, v);
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  pair<int, int> a = dfs(1, 0);
  pair<int, int> b = dfs(a.second, 0);
  a = dfs(b.second, 0);
  queue<int> q;
  q.push(a.second);
  vst[a.second] = 1;
  while (!q.empty()) {
    u = q.front();
    q.pop();
    for (int i = 0; i < adj[u].size(); i++) {
      v = adj[u][i];
      if (vst[v]) continue;
      vst[v] = vst[u] + 1;
      q.push(v);
      pre[v] = u;
    }
  }
  for (int i = b.second; i != a.second; i = pre[i]) vst[i] = -vst[i];
  vst[a.second] = -1;
  for (int i = 1; i <= n; i++) vst[i] < 0 ? vst[i] = -vst[i] : vst[i] = 0;
  for (int i = b.second; i != a.second; i = pre[i]) {
    res.push_back(pair<int, int>(a.second, i));
    ans += vst[i] - vst[a.second];
  }
  reverse(res.begin(), res.end());
  for (int i = pre[b.second]; i != a.second; i = pre[i]) {
    u = vst[b.second] - vst[i];
    v = vst[i] - vst[a.second];
    solve(i, u > v ? pair<int, int>(b.second, u) : pair<int, int>(a.second, v));
  }
  println(ans);
  for (int i = res.size(); i--;)
    print(res[i].first, res[i].second, res[i].second);
  return 0;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
template <typename T, typename U>
inline void smin(T &a, U b) {
  if (a > b) a = b;
}
template <typename T, typename U>
inline void smax(T &a, U b) {
  if (a < b) a = b;
}
template <class T>
inline void gn(T &first) {
  char c, sg = 0;
  while (c = getchar(), (c > '9' || c < '0') && c != '-')
    ;
  for ((c == '-' ? sg = 1, c = getchar() : 0), first = 0; c >= '0' && c <= '9';
       c = getchar())
    first = (first << 1) + (first << 3) + c - '0';
  if (sg) first = -first;
}
template <class T, class T1>
inline void gn(T &first, T1 &second) {
  gn(first);
  gn(second);
}
template <class T, class T1, class T2>
inline void gn(T &first, T1 &second, T2 &z) {
  gn(first);
  gn(second);
  gn(z);
}
template <class T>
inline void print(T first) {
  if (first < 0) {
    putchar('-');
    return print(-first);
  }
  if (first < 10) {
    putchar('0' + first);
    return;
  }
  print(first / 10);
  putchar(first % 10 + '0');
}
template <class T>
inline void printsp(T first) {
  print(first);
  putchar(' ');
}
template <class T>
inline void println(T first) {
  print(first);
  putchar('\n');
}
template <class T, class U>
inline void print(T first, U second) {
  printsp(first);
  println(second);
}
template <class T, class U, class V>
inline void print(T first, U second, V z) {
  printsp(first);
  printsp(second);
  println(z);
}
int power(int a, int b, int m, int ans = 1) {
  for (; b; b >>= 1, a = 1LL * a * a % m)
    if (b & 1) ans = 1LL * ans * a % m;
  return ans;
}
vector<int> adj[200200];
pair<int, int> dfs(int u, int d, int pa = -1) {
  pair<int, int> first = pair<int, int>(d, u);
  for (int i = 0; i < adj[u].size(); i++)
    if (adj[u][i] != pa) smax(first, dfs(adj[u][i], d + 1, u));
  return first;
}
long long ans;
vector<pair<int, int> > res;
int vst[200200], pre[200200];
void solve(int u, pair<int, int> d, int pa = -1) {
  for (int i = 0; i < adj[u].size(); i++) {
    int v = adj[u][i];
    if (v == pa || vst[v]) continue;
    res.push_back(pair<int, int>(d.first, v));
    ans += d.second + 1;
    solve(v, pair<int, int>(d.first, d.second + 1), u);
  }
}
int main() {
  int n, u, v;
  gn(n);
  if (n == 2) {
    puts("1");
    puts("1 2 2");
    return 0;
  }
  for (int i = 1; i < n; i++) {
    gn(u, v);
    if (u > v) swap(u, v);
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  pair<int, int> a = dfs(1, 0);
  pair<int, int> b = dfs(a.second, 0);
  a = dfs(b.second, 0);
  queue<int> q;
  q.push(a.second);
  vst[a.second] = 1;
  while (!q.empty()) {
    u = q.front();
    q.pop();
    for (int i = 0; i < adj[u].size(); i++) {
      v = adj[u][i];
      if (vst[v]) continue;
      vst[v] = vst[u] + 1;
      q.push(v);
      pre[v] = u;
    }
  }
  for (int i = b.second; i != a.second; i = pre[i]) vst[i] = -vst[i];
  vst[a.second] = -1;
  for (int i = 1; i <= n; i++) vst[i] < 0 ? vst[i] = -vst[i] : vst[i] = 0;
  for (int i = b.second; i != a.second; i = pre[i]) {
    res.push_back(pair<int, int>(a.second, i));
    ans += vst[i] - vst[a.second];
  }
  reverse(res.begin(), res.end());
  for (int i = pre[b.second]; i != a.second; i = pre[i]) {
    u = vst[b.second] - vst[i];
    v = vst[i] - vst[a.second];
    solve(i, u > v ? pair<int, int>(b.second, u) : pair<int, int>(a.second, v));
  }
  println(ans);
  for (int i = res.size(); i--;)
    print(res[i].first, res[i].second, res[i].second);
  return 0;
}

```