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;
long long n;
vector<int> g[200020];
int d1[200020], d2[200020];
int deg[200020];
bool vis[200020];
vector<pair<pair<long long, long long>, long long> > resb;
pair<long long, long long> dfs(long long v, long long pre, int d[],
                               long long nd) {
  pair<long long, long long> res = pair<long long, long long>(0, v);
  d[v] = nd;
  for (long long nv : g[v]) {
    if (nv == pre) continue;
    pair<long long, long long> tmp = dfs(nv, v, d, nd + 1);
    tmp.first++;
    res = (res > tmp ? res : tmp);
  }
  return res;
}
int main() {
  cin >> n;
  for (long long i = (long long)(0); i < (long long)(n - 1); i++) {
    long long a, b;
    cin >> a >> b;
    a--;
    b--;
    g[a].push_back(b);
    g[b].push_back(a);
  }
  long long u, v;
  pair<long long, long long> tmp = dfs(0, -1, d1, 0);
  u = tmp.second;
  tmp = dfs(u, -1, d1, 0);
  v = tmp.second;
  dfs(v, -1, d2, 0);
  for (long long i = (long long)(0); i < (long long)(n); i++)
    deg[i] = g[i].size();
  priority_queue<pair<long long, long long> > que;
  for (long long i = (long long)(0); i < (long long)(n); i++) {
    if (i == u || i == v) continue;
    if (deg[i] == 1) {
      que.push(pair<long long, long long>(max(d1[i], d2[i]), i));
    }
  }
  long long resa = 0;
  while (que.size()) {
    pair<long long, long long> crt = que.top();
    que.pop();
    resa += crt.first;
    if (d1[crt.second] > d2[crt.second])
      resb.push_back(pair<pair<long long, long long>, long long>(
          pair<long long, long long>(crt.second + 1, u + 1), crt.second + 1));
    else
      resb.push_back(pair<pair<long long, long long>, long long>(
          pair<long long, long long>(crt.second + 1, v + 1), crt.second + 1));
    vis[crt.second] = true;
    for (long long nv : g[crt.second]) {
      deg[nv]--;
      if (deg[nv] == 1) {
        que.push(pair<long long, long long>(max(d1[nv], d2[nv]), nv));
      }
    }
  }
  long long crt = u;
  while (crt != v) {
    resa += d2[crt];
    resb.push_back(pair<pair<long long, long long>, long long>(
        pair<long long, long long>(crt + 1, v + 1), crt + 1));
    vis[crt] = true;
    for (long long nv : g[crt]) {
      if (!vis[nv]) {
        crt = nv;
        break;
      }
    }
  }
  cout << resa << endl;
  for (long long i = (long long)(0); i < (long long)(resb.size()); i++)
    cout << resb[i].first.first << " " << resb[i].first.second << " "
         << resb[i].second << endl;
  return 0;
}