#include <bits/stdc++.h>
using namespace std;
const int N = 200005;
int d[3][N], n, x, y;
long long ans;
vector<int> e[N], out;
queue<int> q;
void bfs(int *dis, int b) {
  for (int i = 1; i <= n; i++) dis[i] = -1;
  dis[b] = 0;
  q.push(b);
  while (!q.empty()) {
    int u = q.front();
    q.pop();
    for (auto v : e[u]) {
      if (dis[v] == -1) {
        dis[v] = dis[u] + 1;
        q.push(v);
      }
    }
  }
}
int dfs(int u, int f, int t) {
  int zh = 0;
  for (auto v : e[u]) {
    if (v == f) continue;
    zh |= dfs(v, u, t);
  }
  if (u == y) zh = 1;
  if (t == 0 && zh == 0) {
    out.push_back(u);
    out.push_back(d[1][u] > d[2][u] ? x : y);
    out.push_back(u);
    ans += max(d[1][u], d[2][u]);
  }
  if (t == 1 && zh == 1 && f) {
    out.push_back(u);
    out.push_back(x);
    out.push_back(u);
    ans += d[1][u];
  }
  return zh;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i < n; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    e[u].push_back(v);
    e[v].push_back(u);
  }
  bfs(d[0], 1);
  for (int i = 1; i <= n; i++)
    if (d[0][i] > d[0][x]) x = i;
  bfs(d[1], x);
  for (int i = 1; i <= n; i++)
    if (d[1][i] > d[1][y]) y = i;
  bfs(d[2], y);
  dfs(x, 0, 0);
  dfs(x, 0, 1);
  printf("%I64d\n", ans);
  for (int i = 0; i < out.size(); i += 3)
    printf("%d %d %d\n", out[i], out[i + 1], out[i + 2]);
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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;
int d[3][N], n, x, y;
long long ans;
vector<int> e[N], out;
queue<int> q;
void bfs(int *dis, int b) {
  for (int i = 1; i <= n; i++) dis[i] = -1;
  dis[b] = 0;
  q.push(b);
  while (!q.empty()) {
    int u = q.front();
    q.pop();
    for (auto v : e[u]) {
      if (dis[v] == -1) {
        dis[v] = dis[u] + 1;
        q.push(v);
      }
    }
  }
}
int dfs(int u, int f, int t) {
  int zh = 0;
  for (auto v : e[u]) {
    if (v == f) continue;
    zh |= dfs(v, u, t);
  }
  if (u == y) zh = 1;
  if (t == 0 && zh == 0) {
    out.push_back(u);
    out.push_back(d[1][u] > d[2][u] ? x : y);
    out.push_back(u);
    ans += max(d[1][u], d[2][u]);
  }
  if (t == 1 && zh == 1 && f) {
    out.push_back(u);
    out.push_back(x);
    out.push_back(u);
    ans += d[1][u];
  }
  return zh;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i < n; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    e[u].push_back(v);
    e[v].push_back(u);
  }
  bfs(d[0], 1);
  for (int i = 1; i <= n; i++)
    if (d[0][i] > d[0][x]) x = i;
  bfs(d[1], x);
  for (int i = 1; i <= n; i++)
    if (d[1][i] > d[1][y]) y = i;
  bfs(d[2], y);
  dfs(x, 0, 0);
  dfs(x, 0, 1);
  printf("%I64d\n", ans);
  for (int i = 0; i < out.size(); i += 3)
    printf("%d %d %d\n", out[i], out[i + 1], out[i + 2]);
  return 0;
}

```