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;
set<long long> g[500001];
long long dist[500001][2], par[500001];
long long n;
long long bfs(long long a) {
  queue<long long> q;
  vector<long long> dis(500001, 100000001);
  q.push(a);
  dis[a] = 0;
  par[a] = -1;
  while (q.size()) {
    long long u = q.front();
    q.pop();
    for (auto it = g[u].begin(); it != g[u].end(); ++it) {
      long long v = *it;
      if (dis[v] > dis[u] + 1) {
        par[v] = u;
        dis[v] = dis[u] + 1;
        q.push(v);
      }
    }
  }
  long long mx = 0, ans;
  for (long long i = 1; i <= n; i++) {
    if (dis[i] > mx) {
      mx = dis[i];
      ans = i;
    }
  }
  return ans;
}
void _bfs(long long a, bool b) {
  queue<long long> q;
  dist[a][b] = 0;
  q.push(a);
  while (q.size()) {
    long long u = q.front();
    q.pop();
    for (auto it = g[u].begin(); it != g[u].end(); ++it) {
      long long v = *it;
      if (dist[v][b] > dist[u][b] + 1) {
        dist[v][b] = dist[u][b] + 1;
        q.push(v);
      }
    }
  }
}
queue<long long> leaf;
vector<pair<long long, long long> > ans;
long long sum = 0;
void solve(long long u, long long par, long long v, long long d) {
  if (u == v) return;
  ans.push_back({v, u});
  sum += d;
  for (auto it = g[u].begin(); it != g[u].end(); ++it) {
    if (*it != par) {
      solve(*it, u, v, --d);
    }
  }
  return;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  long long a, b;
  for (long long i = 0; i < n - 1; i++) {
    cin >> a >> b;
    g[a].insert(b);
    g[b].insert(a);
  }
  long long u = bfs(1);
  long long v = bfs(u);
  for (long long i = 0; i < 500001; i++) {
    dist[i][0] = 100000001;
    dist[i][1] = 100000001;
  }
  _bfs(u, 0);
  _bfs(v, 1);
  for (long long i = 1; i <= n; i++) {
    if (g[i].size() == 1 && i != v && i != u) {
      leaf.push(i);
    }
  }
  while (leaf.size()) {
    long long x = leaf.front();
    leaf.pop();
    g[par[x]].erase(x);
    g[x].erase(par[x]);
    if (g[par[x]].size() == 1) leaf.push(par[x]);
    if (dist[x][0] > dist[x][1])
      ans.push_back({u, x});
    else
      ans.push_back({v, x});
    sum += max(dist[x][0], dist[x][1]);
  }
  solve(u, -1, v, dist[v][0]);
  cout << sum << endl;
  for (long long i = 0; i < ans.size(); i++) {
    cout << ans[i].first << " " << ans[i].second << " " << ans[i].second
         << endl;
  }
  return 0;
}