#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 7;
const int inf = 0x3f3f3f3f;
const long long INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 + 7;
const double eps = 1e-8;
const double PI = acos(-1);
int n, deg[N], deptha[N], depthb[N];
vector<int> G[N];
queue<int> que;
vector<pair<pair<int, int>, int>> vc;
pair<int, int> dfs(int u, int fa, int* depth) {
  if (fa)
    depth[u] = depth[fa] + 1;
  else
    depth[u] = 0;
  pair<int, int> ans = make_pair(depth[u], u);
  for (auto& v : G[u]) {
    if (v == fa) continue;
    ans = max(ans, dfs(v, u, depth));
  }
  return ans;
}
int main() {
  scanf("%d", &n);
  for (int i = 2; i <= n; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    G[u].push_back(v);
    G[v].push_back(u);
    deg[u]++;
    deg[v]++;
  }
  int A = dfs(1, 0, deptha).second;
  int B = dfs(A, 0, deptha).second;
  dfs(B, 0, depthb);
  for (int i = 1; i <= n; i++)
    if (i != A && i != B && deg[i] == 1) que.push(i);
  long long ans = 0;
  while (!que.empty()) {
    int u = que.front();
    que.pop();
    if (deptha[u] > depthb[u]) {
      ans += deptha[u];
      vc.push_back(make_pair(make_pair(A, u), u));
    } else {
      ans += depthb[u];
      vc.push_back(make_pair(make_pair(B, u), u));
    }
    for (auto& v : G[u])
      if (--deg[v] == 1) que.push(v);
  }
  que.push(B);
  while (!que.empty()) {
    int u = que.front();
    que.pop();
    if (u == A) break;
    ans += deptha[u];
    vc.push_back(make_pair(make_pair(A, u), u));
    for (auto& v : G[u])
      if (--deg[v] == 1) que.push(v);
  }
  printf("%lld\n", ans);
  for (auto& t : vc)
    printf("%d %d %d\n", t.first.first, t.first.second, t.second);
  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 = 2e5 + 7;
const int inf = 0x3f3f3f3f;
const long long INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 + 7;
const double eps = 1e-8;
const double PI = acos(-1);
int n, deg[N], deptha[N], depthb[N];
vector<int> G[N];
queue<int> que;
vector<pair<pair<int, int>, int>> vc;
pair<int, int> dfs(int u, int fa, int* depth) {
  if (fa)
    depth[u] = depth[fa] + 1;
  else
    depth[u] = 0;
  pair<int, int> ans = make_pair(depth[u], u);
  for (auto& v : G[u]) {
    if (v == fa) continue;
    ans = max(ans, dfs(v, u, depth));
  }
  return ans;
}
int main() {
  scanf("%d", &n);
  for (int i = 2; i <= n; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    G[u].push_back(v);
    G[v].push_back(u);
    deg[u]++;
    deg[v]++;
  }
  int A = dfs(1, 0, deptha).second;
  int B = dfs(A, 0, deptha).second;
  dfs(B, 0, depthb);
  for (int i = 1; i <= n; i++)
    if (i != A && i != B && deg[i] == 1) que.push(i);
  long long ans = 0;
  while (!que.empty()) {
    int u = que.front();
    que.pop();
    if (deptha[u] > depthb[u]) {
      ans += deptha[u];
      vc.push_back(make_pair(make_pair(A, u), u));
    } else {
      ans += depthb[u];
      vc.push_back(make_pair(make_pair(B, u), u));
    }
    for (auto& v : G[u])
      if (--deg[v] == 1) que.push(v);
  }
  que.push(B);
  while (!que.empty()) {
    int u = que.front();
    que.pop();
    if (u == A) break;
    ans += deptha[u];
    vc.push_back(make_pair(make_pair(A, u), u));
    for (auto& v : G[u])
      if (--deg[v] == 1) que.push(v);
  }
  printf("%lld\n", ans);
  for (auto& t : vc)
    printf("%d %d %d\n", t.first.first, t.first.second, t.second);
  return 0;
}

```