#include <bits/stdc++.h>
using namespace std;
vector<int> *graph;
int *dep, *dep2;
vector<pair<int, int>> ops;
long long t = 0;
int root, bot;
inline void op(int a, int b, int d) {
  ops.push_back(make_pair(a + 1, b + 1));
  t += d;
}
void dfs1(int u, int p) {
  if (p != -1) dep2[u] = dep2[p] + 1;
  for (int i = 0; i < graph[u].size() - (p != -1); ++i) {
    if (graph[u][i] == p) swap(graph[u][i], graph[u].back());
    int v = graph[u][i];
    dfs1(v, u);
    dep[u] = max(dep[v] + 1, dep[u]);
    if (dep[v] > dep[graph[u][0]]) swap(graph[u][i], graph[u][0]);
  }
}
void dfs2(int u, int p, int d, int d2) {
  if (graph[u].size() - (p != -1)) {
    dfs2(graph[u][0], u, d ? d + 1 : 0, d2);
    for (int i = 1; i < graph[u].size() - (p != -1); ++i)
      dfs2(graph[u][i], u, d ? d + 1 : max(dep[u], dep2[u]) + 1,
           !d && dep[u] > dep2[u] ? bot : d2);
  }
  if (d) op(d2, u, d);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  graph = new vector<int>[n];
  dep = new int[n](), dep2 = new int[n]();
  for (int i = 0; i < n - 1; ++i) {
    int u, v;
    cin >> u >> v;
    --u, --v;
    graph[u].push_back(v);
    graph[v].push_back(u);
  }
  dfs1(0, -1);
  for (; !root || graph[root].size() > 1; root = graph[root][0])
    ;
  memset(dep, 0, 4 * n), memset(dep2, 0, 4 * n);
  dfs1(root, -1);
  for (bot = root; bot == root || graph[bot].size() > 1; bot = graph[bot][0])
    ;
  dfs2(root, -1, 0, root);
  for (; root != bot; root = graph[root][0]) op(bot, root, dep[root]);
  cout << t << "\n";
  for (pair<int, int> p : ops)
    cout << p.first << " " << p.second << " " << p.second << "\n";
}

### Prompt

Please formulate a cpp solution to the following 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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
vector<int> *graph;
int *dep, *dep2;
vector<pair<int, int>> ops;
long long t = 0;
int root, bot;
inline void op(int a, int b, int d) {
  ops.push_back(make_pair(a + 1, b + 1));
  t += d;
}
void dfs1(int u, int p) {
  if (p != -1) dep2[u] = dep2[p] + 1;
  for (int i = 0; i < graph[u].size() - (p != -1); ++i) {
    if (graph[u][i] == p) swap(graph[u][i], graph[u].back());
    int v = graph[u][i];
    dfs1(v, u);
    dep[u] = max(dep[v] + 1, dep[u]);
    if (dep[v] > dep[graph[u][0]]) swap(graph[u][i], graph[u][0]);
  }
}
void dfs2(int u, int p, int d, int d2) {
  if (graph[u].size() - (p != -1)) {
    dfs2(graph[u][0], u, d ? d + 1 : 0, d2);
    for (int i = 1; i < graph[u].size() - (p != -1); ++i)
      dfs2(graph[u][i], u, d ? d + 1 : max(dep[u], dep2[u]) + 1,
           !d && dep[u] > dep2[u] ? bot : d2);
  }
  if (d) op(d2, u, d);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  graph = new vector<int>[n];
  dep = new int[n](), dep2 = new int[n]();
  for (int i = 0; i < n - 1; ++i) {
    int u, v;
    cin >> u >> v;
    --u, --v;
    graph[u].push_back(v);
    graph[v].push_back(u);
  }
  dfs1(0, -1);
  for (; !root || graph[root].size() > 1; root = graph[root][0])
    ;
  memset(dep, 0, 4 * n), memset(dep2, 0, 4 * n);
  dfs1(root, -1);
  for (bot = root; bot == root || graph[bot].size() > 1; bot = graph[bot][0])
    ;
  dfs2(root, -1, 0, root);
  for (; root != bot; root = graph[root][0]) op(bot, root, dep[root]);
  cout << t << "\n";
  for (pair<int, int> p : ops)
    cout << p.first << " " << p.second << " " << p.second << "\n";
}

```