#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 11;
int t, dist, L, R, kol[N], use[N];
int dist1[N], dist2[N];
vector<int> v[N];
void dfs(int l, int r, int pr) {
  if (r > dist) {
    dist = r;
    t = l;
  }
  for (int j = 0; j < v[l].size(); j++)
    if (v[l][j] != pr) dfs(v[l][j], r + 1, l);
}
void dfs2(int l, int ds, int pr, int kk) {
  if (kk == 0)
    dist1[l] = ds;
  else
    dist2[l] = ds;
  for (int j = 0; j < v[l].size(); j++)
    if (v[l][j] != pr) dfs2(v[l][j], ds + 1, l, kk);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  for (int i = 1; i < n; i++) {
    int l, r;
    cin >> l >> r;
    v[l].push_back(r);
    v[r].push_back(l);
  }
  t = 1;
  dist = 0;
  dfs(1, 0, 0);
  dist = 0;
  L = t;
  dfs(t, 0, 0);
  R = t;
  dfs2(L, 0, 0, 0);
  dfs2(R, 0, 0, 1);
  vector<int> vv;
  for (int i = 1; i <= n; i++) kol[i] = v[i].size();
  for (int i = 1; i <= n; i++)
    if (kol[i] == 1) vv.push_back(i);
  long long ans = 0;
  vector<pair<int, int> > res;
  for (int j = 0; j < vv.size(); j++)
    if (vv[j] != L && vv[j] != R) {
      int to = vv[j];
      if (dist1[to] > dist2[to]) {
        ans += 1ll * dist1[to];
        res.push_back(make_pair(L, to));
      } else {
        ans += 1ll * dist2[to];
        res.push_back(make_pair(R, to));
      }
      for (int p = 0; p < v[to].size(); p++) {
        kol[v[to][p]]--;
        if (kol[v[to][p]] == 1) vv.push_back(v[to][p]);
      }
      use[to] = 1;
    }
  while (L != R) {
    ans += 1ll * dist1[R];
    use[R] = 1;
    res.push_back(make_pair(L, R));
    for (int j = 0; j < v[R].size(); j++)
      if (use[v[R][j]] == 0) {
        R = v[R][j];
        break;
      }
  }
  cout << ans << "\n";
  for (int j = 0; j < res.size(); j++)
    cout << res[j].first << " " << res[j].second << " " << res[j].second
         << "\n";
}

### Prompt

Generate 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;
const int N = 2e5 + 11;
int t, dist, L, R, kol[N], use[N];
int dist1[N], dist2[N];
vector<int> v[N];
void dfs(int l, int r, int pr) {
  if (r > dist) {
    dist = r;
    t = l;
  }
  for (int j = 0; j < v[l].size(); j++)
    if (v[l][j] != pr) dfs(v[l][j], r + 1, l);
}
void dfs2(int l, int ds, int pr, int kk) {
  if (kk == 0)
    dist1[l] = ds;
  else
    dist2[l] = ds;
  for (int j = 0; j < v[l].size(); j++)
    if (v[l][j] != pr) dfs2(v[l][j], ds + 1, l, kk);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  for (int i = 1; i < n; i++) {
    int l, r;
    cin >> l >> r;
    v[l].push_back(r);
    v[r].push_back(l);
  }
  t = 1;
  dist = 0;
  dfs(1, 0, 0);
  dist = 0;
  L = t;
  dfs(t, 0, 0);
  R = t;
  dfs2(L, 0, 0, 0);
  dfs2(R, 0, 0, 1);
  vector<int> vv;
  for (int i = 1; i <= n; i++) kol[i] = v[i].size();
  for (int i = 1; i <= n; i++)
    if (kol[i] == 1) vv.push_back(i);
  long long ans = 0;
  vector<pair<int, int> > res;
  for (int j = 0; j < vv.size(); j++)
    if (vv[j] != L && vv[j] != R) {
      int to = vv[j];
      if (dist1[to] > dist2[to]) {
        ans += 1ll * dist1[to];
        res.push_back(make_pair(L, to));
      } else {
        ans += 1ll * dist2[to];
        res.push_back(make_pair(R, to));
      }
      for (int p = 0; p < v[to].size(); p++) {
        kol[v[to][p]]--;
        if (kol[v[to][p]] == 1) vv.push_back(v[to][p]);
      }
      use[to] = 1;
    }
  while (L != R) {
    ans += 1ll * dist1[R];
    use[R] = 1;
    res.push_back(make_pair(L, R));
    for (int j = 0; j < v[R].size(); j++)
      if (use[v[R][j]] == 0) {
        R = v[R][j];
        break;
      }
  }
  cout << ans << "\n";
  for (int j = 0; j < res.size(); j++)
    cout << res[j].first << " " << res[j].second << " " << res[j].second
         << "\n";
}

```