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;
const int N = 200200;
int n;
vector<int> adj[N];
int deg[N], dis[N];
int best, mx, first = -1, second = -1;
void dfs(int v, int p = -1, int cur = 0) {
  dis[v] = cur;
  if (cur > mx) {
    best = v;
    mx = cur;
  }
  for (int u : adj[v]) {
    if (u == p) continue;
    dfs(u, v, cur + 1);
  }
}
void go(int v, int p, int cur, vector<int>& all) {
  all[v] = cur;
  for (int u : adj[v]) {
    if (u == p) continue;
    go(u, v, cur + 1, all);
  }
}
int main() {
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    cin >> a >> b;
    adj[a].push_back(b);
    adj[b].push_back(a);
    deg[a]++;
    deg[b]++;
  }
  dfs(1);
  mx = -1;
  int v = best;
  dfs(v);
  mx = best;
  vector<int> dis_v(n + 1), dis_u(n + 1);
  go(v, -1, 0, dis_v);
  go(mx, -1, 0, dis_u);
  long long int ans = 0;
  vector<vector<int>> an;
  auto add = [&](int a, int b, int c) { an.push_back({a, b, c}); };
  vector<int> all;
  vector<bool> mark(n + 1);
  for (int i = 1; i <= n; i++) {
    if (i == v || i == mx) continue;
    if (deg[i] == 1) {
      all.push_back(i);
    }
  }
  int cnt = 0;
  for (int i = 0; i < (int)all.size(); i++) {
    int it = all[i];
    cnt++;
    mark[it] = 1;
    if (dis_v[it] > dis_u[it]) {
      ans += dis_v[it];
      add(it, v, it);
    } else {
      ans += dis_u[it];
      add(it, mx, it);
    }
    for (int u : adj[it]) {
      deg[u]--;
      if (deg[u] == 1) all.push_back(u);
    }
  }
  assert(n - cnt == dis_v[mx] + 1);
  all.clear();
  for (int i = 1; i <= n; i++) {
    if (i != v && !mark[i] && deg[i] == 1) {
      all.push_back(i);
    }
  }
  for (int i = 0; i < (int)all.size(); i++) {
    ans += dis_v[all[i]];
    cnt++;
    add(v, all[i], all[i]);
    for (int u : adj[all[i]]) {
      deg[u]--;
      if (deg[u] == 1) all.push_back(u);
    }
  }
  assert(cnt == n - 1);
  cout << ans << '\n';
  for (auto it : an) {
    cout << it[0] << ' ' << it[1] << ' ' << it[2] << '\n';
  }
  return 0;
}