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 max_n = 200005;
int n;
vector<int> G[max_n];
int in[max_n];
int a, b;
int par[max_n];
int d[2][max_n];
vector<int> td;
vector<pair<int, int> > res;
void dfs(int u, int fa, int d[]) {
  if (fa != -1) par[u] = fa;
  for (auto v : G[u]) {
    if (v == fa) continue;
    d[v] = d[u] + 1;
    dfs(v, u, d);
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i < n; i++) {
    scanf("%d%d", &a, &b);
    G[a].push_back(b);
    G[b].push_back(a);
    in[a]++;
    in[b]++;
  }
  d[0][1] = 0;
  dfs(1, -1, d[0]);
  int mv = 0;
  for (int i = 1; i <= n; i++) mv = max(mv, d[0][i]);
  int l, r;
  for (int i = 1; i <= n; i++) {
    if (d[0][i] == mv) {
      l = i;
      break;
    }
  }
  d[0][l] = 0;
  dfs(l, -1, d[0]);
  mv = 0;
  for (int i = 1; i <= n; i++) mv = max(mv, d[0][i]);
  for (int i = 1; i <= n; i++) {
    if (mv == d[0][i]) {
      r = i;
      break;
    }
  }
  int now = r;
  while (1) {
    td.push_back(now);
    if (now == l) break;
    now = par[now];
  }
  dfs(r, -1, d[1]);
  queue<int> q;
  for (int i = 1; i <= n; i++)
    if (in[i] == 1 && i != l && i != r) q.push(i);
  long long ans = 0;
  int p;
  while (!q.empty()) {
    p = q.front();
    q.pop();
    if (d[0][p] <= d[1][p])
      ans += d[1][p], res.push_back(pair<int, int>(p, r));
    else
      ans += d[0][p], res.push_back(pair<int, int>(p, l));
    if (par[p] == l || par[p] == r) continue;
    in[par[p]]--;
    if (in[par[p]] == 1) q.push(par[p]);
  }
  for (int i = td.size() - 1; i >= 1; i--)
    ans += d[1][td[i]], res.push_back(pair<int, int>(td[i], r));
  printf("%lld\n", ans);
  for (auto p : res) printf("%d %d %d\n", p.first, p.second, p.first);
  return 0;
}