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 long long mod = 1e9 + 7, siz = 1e6 + 5;
vector<long long> g[siz];
long long n, t, m, k, a[siz], d1[siz], d2[siz];
set<pair<long long, long long>, greater<pair<long long, long long> > > s;
bool vis[siz];
bool cmp(long long a, long long b) { return d1[a] < d1[b]; }
void go(long long u, long long p, long long *dist) {
  for (auto v : g[u]) {
    if (v == p) continue;
    dist[v] = 1 + dist[u];
    go(v, u, dist);
  }
}
void find_dep(long long u, long long lvl) {
  vis[u] = 1;
  for (auto v : g[u]) {
    if (vis[v]) continue;
    s.insert({lvl + 1, v});
    find_dep(v, lvl + 1);
  }
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (long long i = 1; i < n; i++) {
    long long x, y;
    cin >> x >> y;
    g[x].push_back(y);
    g[y].push_back(x);
  }
  go(1, 0, d1);
  long long side1, side2, sofar = -1;
  for (long long i = 1; i <= n; i++) {
    if (d1[i] > sofar) {
      sofar = d1[i];
      side1 = i;
    }
  }
  go(side1, 0, d2);
  sofar = -1;
  for (long long i = 1; i <= n; i++) {
    if (d2[i] > sofar) {
      sofar = d2[i];
      side2 = i;
    }
  }
  long long dia = d2[side2];
  d1[side2] = 0;
  go(side2, 0, d1);
  long long ans = 0;
  vector<pair<pair<long long, long long>, long long> > out;
  vector<long long> chain;
  for (long long i = 1; i <= n; i++)
    if (d1[i] + d2[i] == dia) {
      chain.push_back(i);
      vis[i] = 1;
    }
  sort(chain.begin(), chain.end(), cmp);
  for (auto x : chain) {
    find_dep(x, 0);
    if (d2[x] >= d1[x]) {
      while (s.size()) {
        pair<long long, long long> p = *s.begin();
        long long node = p.second;
        ans += d2[node];
        out.push_back({{side1, node}, node});
        s.erase(s.begin());
      }
    } else {
      while (s.size()) {
        pair<long long, long long> p = *s.begin();
        long long node = p.second;
        ans += d1[node];
        out.push_back({{side2, node}, node});
        s.erase(s.begin());
      }
    }
  }
  long long u = chain[0];
  while (chain.size() > 1) {
    ans += dia;
    dia--;
    out.push_back({{u, chain.back()}, chain.back()});
    chain.pop_back();
  }
  cout << ans << endl;
  for (auto p : out)
    cout << p.first.first << " " << p.first.second << " " << p.second << endl;
}