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;
constexpr int N = 2e5 + 10;
long long n, ans, u, v, dist[2][N], h[N], p, q, dia, par[N], d[N];
vector<int> node[N];
queue<int> qu;
bool inpath[N];
void dfs(int v, int p = -1) {
  for (auto i : node[v]) {
    if (i != p) {
      h[i] = h[v] + 1;
      par[i] = v;
      dfs(i, v);
    }
  }
}
void dfs2(int v, int id, int p = -1) {
  for (auto i : node[v]) {
    if (i != p) {
      dist[id][i] = dist[id][v] + 1;
      dfs2(i, id, v);
    }
  }
}
void dfs3(int v, int p = -1) {
  if (v == q) return;
  for (auto i : node[v]) {
    if (i != p && inpath[i]) {
      cout << v << ' ' << q << ' ' << v << '\n';
      dfs3(i, v);
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i < n; i++) {
    cin >> u >> v;
    node[u].push_back(v);
    node[v].push_back(u);
    d[v]++;
    d[u]++;
  }
  dfs(1);
  for (int i = 1; i <= n; i++)
    if (h[i] > h[p]) p = i;
  dfs2(p, 0);
  for (int i = 1; i <= n; i++)
    if (dist[0][i] > dist[0][q]) q = i;
  dfs2(q, 1);
  dia = dist[0][q] + 1;
  u = p;
  v = q;
  if (h[v] < h[u]) swap(v, u);
  while (h[u] < h[v]) {
    inpath[v] = true;
    v = par[v];
  }
  while (v != u) {
    inpath[v] = true;
    inpath[u] = true;
    v = par[v];
    u = par[u];
  }
  inpath[v] = true;
  ans += ((dia - 1) * dia) / 2;
  for (int i = 1; i <= n; i++)
    if (!inpath[i]) ans += max(dist[0][i], dist[1][i]);
  cout << ans << '\n';
  for (int i = 1; i <= n; i++) {
    if (d[i] == 1 && i != p && i != q) qu.push(i);
  }
  while (!qu.empty()) {
    v = qu.front();
    qu.pop();
    if (dist[0][v] > dist[1][v])
      cout << v << ' ' << p << ' ' << v << '\n';
    else
      cout << v << ' ' << q << ' ' << v << '\n';
    for (auto i : node[v]) {
      d[i]--;
      if (d[i] == 1 && i != q && i != p) qu.push(i);
    }
  }
  dfs3(p);
  return 0;
}