#include <bits/stdc++.h>
using namespace std;
int n;
vector<int> e[200001];
int dep[200001], fa[200001], isU[200001];
int m_dist, m_l, root;
long long int ans;
struct Node {
  int a, b, c;
};
vector<Node> op;
void DFS(int x, int fx) {
  fa[x] = fx;
  if (x != fx) dep[x] = dep[fx] + 1;
  if (dep[x] > m_dist) m_dist = dep[x], m_l = x;
  for (auto i : e[x]) {
    if (i != fx) DFS(i, x);
  }
  return;
}
void Destruct(int x, int dist, int rx, int flag = 0) {
  isU[x] = 1;
  for (auto i : e[x]) {
    if (!isU[i]) Destruct(i, dist + 1, rx, 1);
  }
  if (flag) {
    ans += dist;
    op.push_back(Node{rx, x, x});
  }
  return;
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    scanf("%d %d", &a, &b);
    e[a].push_back(b);
    e[b].push_back(a);
  }
  DFS(1, 1);
  m_dist = 0, dep[root = m_l] = 0;
  DFS(root, root);
  fa[root] = 0;
  for (int i = m_l; i; i = fa[i]) isU[i] = 1;
  for (int i = m_l; i; i = fa[i]) {
    if (dep[i] > dep[m_l] - dep[i])
      Destruct(i, dep[i], root);
    else
      Destruct(i, dep[m_l] - dep[i], m_l);
  }
  ans += (long long int)(m_dist + 1) * m_dist / 2;
  printf("%lld\n", ans);
  for (int i = 0; i < op.size(); i++)
    printf("%d %d %d\n", op[i].a, op[i].b, op[i].c);
  for (int i = m_l; i != root; i = fa[i]) printf("%d %d %d\n", root, i, i);
  return 0;
}

### Prompt

Create a solution in CPP for 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;
int n;
vector<int> e[200001];
int dep[200001], fa[200001], isU[200001];
int m_dist, m_l, root;
long long int ans;
struct Node {
  int a, b, c;
};
vector<Node> op;
void DFS(int x, int fx) {
  fa[x] = fx;
  if (x != fx) dep[x] = dep[fx] + 1;
  if (dep[x] > m_dist) m_dist = dep[x], m_l = x;
  for (auto i : e[x]) {
    if (i != fx) DFS(i, x);
  }
  return;
}
void Destruct(int x, int dist, int rx, int flag = 0) {
  isU[x] = 1;
  for (auto i : e[x]) {
    if (!isU[i]) Destruct(i, dist + 1, rx, 1);
  }
  if (flag) {
    ans += dist;
    op.push_back(Node{rx, x, x});
  }
  return;
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    scanf("%d %d", &a, &b);
    e[a].push_back(b);
    e[b].push_back(a);
  }
  DFS(1, 1);
  m_dist = 0, dep[root = m_l] = 0;
  DFS(root, root);
  fa[root] = 0;
  for (int i = m_l; i; i = fa[i]) isU[i] = 1;
  for (int i = m_l; i; i = fa[i]) {
    if (dep[i] > dep[m_l] - dep[i])
      Destruct(i, dep[i], root);
    else
      Destruct(i, dep[m_l] - dep[i], m_l);
  }
  ans += (long long int)(m_dist + 1) * m_dist / 2;
  printf("%lld\n", ans);
  for (int i = 0; i < op.size(); i++)
    printf("%d %d %d\n", op[i].a, op[i].b, op[i].c);
  for (int i = m_l; i != root; i = fa[i]) printf("%d %d %d\n", root, i, i);
  return 0;
}

```