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 ll = long long;
using ld = long double;
using ull = unsigned long long;
using namespace std;
const int N = 2e5 + 2;
int n, d[2][N], par[N], leaf[2], deg[N];
vector<int> g[N];
struct Query {
  int u, v, w;
};
vector<Query> res;
int bfs(int t, int r) {
  queue<int> q;
  for (int i = (1); i <= (n); ++i) d[t][i] = -1;
  d[t][r] = 0;
  q.push(r);
  int lst = 0;
  while (!q.empty()) {
    int u = q.front();
    lst = u;
    q.pop();
    for (int v : g[u])
      if (d[t][v] == -1) {
        d[t][v] = d[t][u] + 1;
        par[v] = u;
        q.push(v);
      }
  }
  return lst;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cin >> n;
  for (int i = (2); i <= (n); ++i) {
    int u, v;
    cin >> u >> v;
    g[u].emplace_back(v);
    g[v].emplace_back(u);
    deg[u]++;
    deg[v]++;
  }
  leaf[0] = bfs(0, 1);
  leaf[1] = bfs(0, leaf[0]);
  bfs(1, leaf[1]);
  queue<int> leaves;
  for (int i = (1); i <= (n); ++i)
    if (deg[i] == 1 && i != leaf[0] && i != leaf[1]) leaves.push(i);
  ll ans = 0;
  while (!leaves.empty()) {
    int u = leaves.front();
    leaves.pop();
    if (--deg[par[u]] == 1) leaves.push(par[u]);
    if (d[0][u] > d[1][u])
      res.push_back({u, leaf[0], u});
    else
      res.push_back({u, leaf[1], u});
    ans += max(d[0][u], d[1][u]);
  }
  int u = leaf[0];
  while (u != leaf[1]) {
    res.push_back({u, leaf[1], u});
    ans += d[1][u];
    u = par[u];
  }
  cout << ans << '\n';
  for (Query Q : res) cout << Q.u << ' ' << Q.v << ' ' << Q.w << '\n';
  return 0;
}