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>
#pragma GCC optimize("Ofast")
using namespace std;
const long long N = 200005;
const long long inf = (long long)1e9;
const long long mod = (long long)1e9 + 7;
clock_t t1, t2;
long long d[N], d1[N], d2[N], n, leaf[N], p1, p2;
bool ok[N];
vector<long long> g[N];
queue<long long> q;
void dfs(long long u, long long pa, long long *x) {
  for (auto i : g[u]) {
    if (i != pa) {
      x[i] = x[u] + 1;
      dfs(i, u, x);
    }
  }
}
void out(long long u, long long pa, long long p) {
  for (auto i : g[u]) {
    if (ok[i] || i == pa) continue;
    cout << u << " " << p << " " << u << '\n';
    out(i, u, p);
  }
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (long long i = 1; i < n; i++) {
    long long u, v;
    cin >> u >> v;
    g[u].push_back(v);
    g[v].push_back(u);
    leaf[u]++;
    leaf[v]++;
  }
  d[1] = 0;
  dfs(1, -1, d);
  long long mx = -1;
  for (long long i = 1; i <= n; i++) {
    if (d[i] > mx) {
      mx = d[i];
      p1 = i;
    }
  }
  d1[p1] = 0;
  dfs(p1, -1, d1);
  mx = -1;
  for (long long i = 1; i <= n; i++) {
    if (d1[i] > mx) {
      mx = d1[i];
      p2 = i;
    }
  }
  d2[p2] = 0;
  dfs(p2, -1, d2);
  for (long long i = 1; i <= n; i++) {
    if (d1[i] + d2[i] != d1[p2] && leaf[i] == 1) {
      q.push(i);
    }
  }
  long long ans = 0, cnt = 0, a[N], b[N], c[N];
  memset(ok, false, sizeof ok);
  while (!q.empty()) {
    long long i = q.front();
    q.pop();
    for (auto u : g[i]) {
      if (ok[u]) continue;
      ans += max(d1[i], d2[i]);
      a[++cnt] = d1[i] > d2[i] ? p1 : p2;
      b[cnt] = i;
      c[cnt] = i;
      leaf[u]--;
      ok[i] = 1;
      if (leaf[u] == 1) q.push(u);
    }
  }
  ans += ((d1[p2] + 1) * (d1[p2])) / 2;
  cout << ans << '\n';
  for (long long i = 1; i <= cnt; i++)
    cout << a[i] << " " << b[i] << " " << c[i] << '\n';
  out(p1, -1, p2);
  return 0;
}