#include <bits/stdc++.h>
using namespace std;
const int N = 200005;
vector<int> g[N];
int ds[N], de[N], diam[N], par[N];
int vis[N];
void dfs(int v, int p, int d, int dist[]) {
  dist[v] = d;
  par[v] = p;
  for (int u : g[v]) {
    if (u != p) dfs(u, v, d + 1, dist);
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cout << setprecision(32);
  int n;
  cin >> n;
  for (int i = 1; i < n; i++) {
    int x, y;
    cin >> x >> y;
    g[x].push_back(y);
    g[y].push_back(x);
  }
  dfs(1, 0, 1, ds);
  int s = 1;
  for (int i = 1; i <= n; i++) {
    if (ds[i] > ds[s]) s = i;
  }
  dfs(s, 0, 1, ds);
  int e = s;
  for (int i = 1; i <= n; i++) {
    if (ds[i] > ds[e]) e = i;
  }
  int v = e;
  vector<int> vec;
  while (v) {
    vec.push_back(v);
    diam[v] = 1;
    v = par[v];
  }
  dfs(e, 0, 1, de);
  vector<pair<int, pair<int, char> > > iic;
  for (int i = 1; i <= n; i++) {
    if (!diam[i]) {
      iic.push_back({ds[i], {i, 's'}});
      iic.push_back({de[i], {i, 'e'}});
    }
  }
  sort(iic.rbegin(), iic.rend());
  memset(vis, 0, sizeof(vis));
  vector<pair<pair<int, int>, int> > ans;
  long long sum = 0;
  for (auto pp : iic) {
    int ind = pp.second.first;
    if (vis[ind]) continue;
    vis[ind] = 1;
    sum += pp.first - 1;
    if (pp.second.second == 's') {
      ans.push_back({{ind, s}, ind});
    } else {
      ans.push_back({{ind, e}, ind});
    }
  }
  sum += vec.size() * 1LL * (vec.size() - 1) / 2;
  for (int i = 0; i + 1 < vec.size(); i++) {
    ans.push_back({{vec[i], vec.back()}, vec[i]});
  }
  cout << sum << '\n';
  for (auto ppp : ans) {
    cout << ppp.first.first << " " << ppp.first.second << " " << ppp.second
         << '\n';
  }
  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;
const int N = 200005;
vector<int> g[N];
int ds[N], de[N], diam[N], par[N];
int vis[N];
void dfs(int v, int p, int d, int dist[]) {
  dist[v] = d;
  par[v] = p;
  for (int u : g[v]) {
    if (u != p) dfs(u, v, d + 1, dist);
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cout << setprecision(32);
  int n;
  cin >> n;
  for (int i = 1; i < n; i++) {
    int x, y;
    cin >> x >> y;
    g[x].push_back(y);
    g[y].push_back(x);
  }
  dfs(1, 0, 1, ds);
  int s = 1;
  for (int i = 1; i <= n; i++) {
    if (ds[i] > ds[s]) s = i;
  }
  dfs(s, 0, 1, ds);
  int e = s;
  for (int i = 1; i <= n; i++) {
    if (ds[i] > ds[e]) e = i;
  }
  int v = e;
  vector<int> vec;
  while (v) {
    vec.push_back(v);
    diam[v] = 1;
    v = par[v];
  }
  dfs(e, 0, 1, de);
  vector<pair<int, pair<int, char> > > iic;
  for (int i = 1; i <= n; i++) {
    if (!diam[i]) {
      iic.push_back({ds[i], {i, 's'}});
      iic.push_back({de[i], {i, 'e'}});
    }
  }
  sort(iic.rbegin(), iic.rend());
  memset(vis, 0, sizeof(vis));
  vector<pair<pair<int, int>, int> > ans;
  long long sum = 0;
  for (auto pp : iic) {
    int ind = pp.second.first;
    if (vis[ind]) continue;
    vis[ind] = 1;
    sum += pp.first - 1;
    if (pp.second.second == 's') {
      ans.push_back({{ind, s}, ind});
    } else {
      ans.push_back({{ind, e}, ind});
    }
  }
  sum += vec.size() * 1LL * (vec.size() - 1) / 2;
  for (int i = 0; i + 1 < vec.size(); i++) {
    ans.push_back({{vec[i], vec.back()}, vec[i]});
  }
  cout << sum << '\n';
  for (auto ppp : ans) {
    cout << ppp.first.first << " " << ppp.first.second << " " << ppp.second
         << '\n';
  }
  return 0;
}

```