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;
template <class T>
struct rge {
  T b, e;
};
template <class T>
rge<T> range(T i, T j) {
  return rge<T>{i, j};
}
struct debug {
  template <class T>
  debug& operator<<(const T&) {
    return *this;
  }
};
using ll = long long;
template <typename T>
void min_self(T& a, T b) {
  a = min(a, b);
}
template <typename T>
void max_self(T& a, T b) {
  a = max(a, b);
}
const int B = 20;
const int mxN = 200 * 1000 + 5;
int par[B][mxN], depth[mxN];
vector<int> g[mxN];
void dfs(int u, int p, int d = 0) {
  par[0][u] = p;
  depth[u] = d;
  for (int b = 1; b < B; b++) {
    par[b][u] = par[b - 1][par[b - 1][u]];
  }
  for (int v : g[u]) {
    if (v != p) {
      dfs(v, u, d + 1);
    }
  }
}
int jump(int u, int k) {
  for (int b = 0; b < B; b++) {
    if (k >> b & 1) {
      u = par[b][u];
    }
  }
  return u;
}
int lca(int u, int v) {
  if (depth[u] > depth[v]) {
    swap(u, v);
  }
  v = jump(v, depth[v] - depth[u]);
  if (u == v) {
    return u;
  }
  for (int b = B - 1; b >= 0; --b) {
    if (par[b][u] != par[b][v]) {
      u = par[b][u];
      v = par[b][v];
    }
  }
  return par[0][u];
}
int dist(int u, int v) { return depth[u] + depth[v] - 2 * depth[lca(u, v)]; }
struct ev {
  int a, b, c;
  void pr() const { cout << a + 1 << ' ' << b + 1 << ' ' << c + 1 << '\n'; }
};
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    cin >> u >> v;
    u--, v--;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  dfs(0, 0);
  auto Bfs = [&](int u) {
    vector<bool> seen(n);
    seen[u] = true;
    vector<int> q{u};
    for (int i = 0; i < (int)q.size(); i++) {
      u = q[i];
      for (int v : g[u]) {
        if (!seen[v]) {
          seen[v] = true;
          q.push_back(v);
        }
      }
    }
    return u;
  };
  int a = Bfs(0);
  int b = Bfs(a);
  vector<int> leaves;
  vector<int> deg(n);
  for (int i = 0; i < n; i++) {
    if ((int)g[i].size() == 1 && i != a && i != b) {
      leaves.push_back(i);
    }
    deg[i] = (int)g[i].size();
  }
  ll score = 0;
  vector<ev> ans;
  for (int rep = 0; rep < 2; rep++) {
    for (int i = 0; i < (int)leaves.size(); i++) {
      int u = leaves[i];
      int da = dist(u, a), db = rep == 1 ? 0 : dist(u, b);
      score += max(da, db);
      if (da >= db) {
        ans.push_back({u, a, u});
      } else {
        ans.push_back({u, b, u});
      }
      for (int v : g[u]) {
        if (--deg[v] == 1) {
          leaves.push_back(v);
        }
      }
    }
    leaves.clear();
    leaves.push_back(b);
  }
  assert((int)ans.size() == n - 1);
  cout << score << '\n';
  for (ev e : ans) {
    e.pr();
  }
}