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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
void dfs(const vector<vector<int>>& g, int a, vector<int>& d, int p) {
  if (a == p) {
    d[a] = 0;
  } else {
    d[a] = d[p] + 1;
  }
  for (int b : g[a]) {
    if (b != p) dfs(g, b, d, a);
  }
}
int main() {
  ios::sync_with_stdio(false);
  int n;
  cin >> n;
  vector<vector<int>> g(n);
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    cin >> a >> b;
    a--;
    b--;
    g[a].push_back(b);
    g[b].push_back(a);
  }
  vector<int> dfb(n);
  dfs(g, 0, dfb, 0);
  int a = 0;
  for (int i = 1; i < n; i++)
    if (dfb[i] > dfb[a]) a = i;
  vector<int> dfa(n);
  dfs(g, a, dfa, a);
  int b = 0;
  for (int i = 1; i < n; i++)
    if (dfa[i] > dfa[b]) b = i;
  dfs(g, b, dfb, b);
  ll val = 0;
  vector<pair<int, int>> ans;
  vector<int> ord;
  for (int i = 0; i < n; i++) {
    if (dfa[i] + dfb[i] == dfa[b]) continue;
    ord.push_back(i);
  }
  sort(ord.begin(), ord.end(),
       [&](int i, int j) { return max(dfa[i], dfb[i]) > max(dfa[j], dfb[j]); });
  for (int i : ord) {
    if (dfa[i] > dfb[i]) {
      val += dfa[i];
      ans.push_back({i, a});
    } else {
      val += dfb[i];
      ans.push_back({i, b});
    }
  }
  ord.clear();
  for (int i = 0; i < n; i++) {
    if (dfa[i] + dfb[i] != dfa[b]) continue;
    ord.push_back(i);
  }
  sort(ord.begin(), ord.end(), [&](int i, int j) { return dfa[i] > dfa[j]; });
  for (int i = 0; i + 1 < ord.size(); i++) {
    val += dfa[ord[i]];
    ans.push_back({ord[i], a});
  }
  cout << val << endl;
  for (auto e : ans) {
    cout << (e.first + 1) << ' ' << (e.second + 1) << ' ' << (e.first + 1)
         << '\n';
  }
}