#include <bits/stdc++.h>
using namespace std;
vector<int> G[202000];
struct P {
  int x, y;
};
vector<P> aa;
int n, x, y, mx, z, rt, fa[202000], dep[202000], in[202000];
long long ans;
void dfs(int x, int fff) {
  dep[x] = dep[fa[x] = fff] + 1;
  if (dep[x] > mx) mx = dep[x], z = x;
  for (auto y : G[x])
    if (y - fff) dfs(y, x);
}
void D(int x, int dis, int fr, int fl = 0) {
  in[x] = 1;
  for (auto y : G[x])
    if (!in[y]) D(y, dis + 1, fr, 1);
  if (fl) ans += dis, aa.push_back(P{fr, x});
}
int main() {
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i < n; ++i)
    cin >> x >> y, G[x].push_back(y), G[y].push_back(x);
  mx = 0;
  dfs(1, 0);
  mx = 0;
  rt = z;
  dfs(rt, 0);
  ans = (long long)mx * (mx - 1) / 2;
  for (int t = z; t; t = fa[t]) in[t] = 1;
  for (int t = z; t; t = fa[t]) {
    if (dep[t] - 1 > dep[z] - dep[t])
      D(t, dep[t] - 1, rt);
    else
      D(t, dep[z] - dep[t], z);
  }
  cout << ans << endl;
  for (int i = 0; i < aa.size(); ++i)
    printf("%d %d %d\n", aa[i].x, aa[i].y, aa[i].y);
  for (int t = z; t - rt; t = fa[t]) printf("%d %d %d\n", rt, t, t);
}

### 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;
vector<int> G[202000];
struct P {
  int x, y;
};
vector<P> aa;
int n, x, y, mx, z, rt, fa[202000], dep[202000], in[202000];
long long ans;
void dfs(int x, int fff) {
  dep[x] = dep[fa[x] = fff] + 1;
  if (dep[x] > mx) mx = dep[x], z = x;
  for (auto y : G[x])
    if (y - fff) dfs(y, x);
}
void D(int x, int dis, int fr, int fl = 0) {
  in[x] = 1;
  for (auto y : G[x])
    if (!in[y]) D(y, dis + 1, fr, 1);
  if (fl) ans += dis, aa.push_back(P{fr, x});
}
int main() {
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i < n; ++i)
    cin >> x >> y, G[x].push_back(y), G[y].push_back(x);
  mx = 0;
  dfs(1, 0);
  mx = 0;
  rt = z;
  dfs(rt, 0);
  ans = (long long)mx * (mx - 1) / 2;
  for (int t = z; t; t = fa[t]) in[t] = 1;
  for (int t = z; t; t = fa[t]) {
    if (dep[t] - 1 > dep[z] - dep[t])
      D(t, dep[t] - 1, rt);
    else
      D(t, dep[z] - dep[t], z);
  }
  cout << ans << endl;
  for (int i = 0; i < aa.size(); ++i)
    printf("%d %d %d\n", aa[i].x, aa[i].y, aa[i].y);
  for (int t = z; t - rt; t = fa[t]) printf("%d %d %d\n", rt, t, t);
}

```