#include <bits/stdc++.h>
#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math,O3")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
const long long N = 2e5 + 10;
const long long big = 1e14;
const long long block = 500;
const long long mod = 1e9 + 7;
long long n;
long long gl[N], gl2[N];
vector<long long> g[N];
pair<long long, long long> mx;
long long dia, z, y, subs;
void orn(long long v, long long p) {
  gl[v] = gl[p] + 1;
  mx = max(mx, {gl[v], v});
  for (auto u : g[v])
    if (u != p) orn(u, v);
}
void orn2(long long v, long long p) {
  gl2[v] = gl2[p] + 1;
  mx = max(mx, {gl2[v], v});
  for (auto u : g[v])
    if (u != p) orn2(u, v);
}
vector<pair<pair<long long, long long>, long long> > ans;
void dfs(long long v, long long p) {
  for (auto u : g[v])
    if (u != p) dfs(u, v);
  if (gl[v] > gl2[v])
    ans.push_back((pair<pair<long long, long long>, long long>){{z, v}, v}),
        subs += gl[v] - 1;
  else
    ans.push_back((pair<pair<long long, long long>, long long>){{y, v}, v}),
        subs += gl2[v] - 1;
}
vector<long long> way;
void sub(long long v, long long p) {
  for (auto u : g[v]) {
    if (u == p) continue;
    if (gl[u] + gl2[u] - 2 == dia)
      sub(u, v);
    else
      dfs(u, v);
  }
  way.push_back(v);
}
void dai() {
  orn(1, 0);
  z = mx.second;
  mx = {0, 0};
  orn(z, 0);
  y = mx.second;
  dia = gl[y] - 1;
  orn2(y, 0);
  sub(y, 0);
  for (auto u : way) {
    if (u == y) break;
    subs += gl2[u] - 1;
    ans.push_back(make_pair(make_pair(y, u), u));
  }
  cout << subs << '\n';
  for (auto u : ans)
    cout << u.first.first << " " << u.first.second << ' ' << u.second << '\n';
}
int main() {
  cin >> n;
  for (int i = 1, a, b; i < n; i++) {
    cin >> a >> b;
    g[a].push_back(b), g[b].push_back(a);
  }
  dai();
}

### Prompt

Develop a solution in CPP 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>
#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math,O3")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
const long long N = 2e5 + 10;
const long long big = 1e14;
const long long block = 500;
const long long mod = 1e9 + 7;
long long n;
long long gl[N], gl2[N];
vector<long long> g[N];
pair<long long, long long> mx;
long long dia, z, y, subs;
void orn(long long v, long long p) {
  gl[v] = gl[p] + 1;
  mx = max(mx, {gl[v], v});
  for (auto u : g[v])
    if (u != p) orn(u, v);
}
void orn2(long long v, long long p) {
  gl2[v] = gl2[p] + 1;
  mx = max(mx, {gl2[v], v});
  for (auto u : g[v])
    if (u != p) orn2(u, v);
}
vector<pair<pair<long long, long long>, long long> > ans;
void dfs(long long v, long long p) {
  for (auto u : g[v])
    if (u != p) dfs(u, v);
  if (gl[v] > gl2[v])
    ans.push_back((pair<pair<long long, long long>, long long>){{z, v}, v}),
        subs += gl[v] - 1;
  else
    ans.push_back((pair<pair<long long, long long>, long long>){{y, v}, v}),
        subs += gl2[v] - 1;
}
vector<long long> way;
void sub(long long v, long long p) {
  for (auto u : g[v]) {
    if (u == p) continue;
    if (gl[u] + gl2[u] - 2 == dia)
      sub(u, v);
    else
      dfs(u, v);
  }
  way.push_back(v);
}
void dai() {
  orn(1, 0);
  z = mx.second;
  mx = {0, 0};
  orn(z, 0);
  y = mx.second;
  dia = gl[y] - 1;
  orn2(y, 0);
  sub(y, 0);
  for (auto u : way) {
    if (u == y) break;
    subs += gl2[u] - 1;
    ans.push_back(make_pair(make_pair(y, u), u));
  }
  cout << subs << '\n';
  for (auto u : ans)
    cout << u.first.first << " " << u.first.second << ' ' << u.second << '\n';
}
int main() {
  cin >> n;
  for (int i = 1, a, b; i < n; i++) {
    cin >> a >> b;
    g[a].push_back(b), g[b].push_back(a);
  }
  dai();
}

```