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;
inline void read(int &x) {
  register int c = getchar();
  x = 0;
  int neg = 0;
  for (; ((c < 48 || c > 57) && c != '-'); c = getchar())
    ;
  if (c == '-') {
    neg = 1;
    c = getchar();
  }
  for (; c > 47 && c < 58; c = getchar()) {
    x = (x << 1) + (x << 3) + c - 48;
  }
  if (neg) x = -x;
}
inline void writeln(int x) {
  char buffor[21];
  register int i = 0;
  int neg = 0;
  if (x < 0) {
    neg = 1;
    x = -x;
  }
  do {
    buffor[i++] = (x % 10) + '0';
    x /= 10;
  } while (x);
  i--;
  if (neg) putchar('-');
  while (i >= 0) putchar(buffor[i--]);
  putchar('\n');
}
const int N = 2e5 + 10;
int n;
vector<int> a[N];
int d[N], par[N], inpath[N], h[N], maxlen, trace[N];
long long res = 0;
struct Ans {
  int a, b, c;
} ans[N];
int cnt = 0, s, t;
void input() {
  cin >> n;
  for (int i = 1; i <= n - 1; i++) {
    int u, v;
    u = i;
    cin >> u >> v;
    a[u].push_back(v);
    a[v].push_back(u);
  }
}
int bfs(int s) {
  for (int i = 1; i <= n; i++) d[i] = -1;
  d[s] = 0;
  trace[s] = s;
  queue<int> q;
  q.push(s);
  while (!q.empty()) {
    int u = q.front();
    q.pop();
    for (auto v : a[u])
      if (d[v] == -1) {
        d[v] = d[u] + 1;
        trace[v] = u;
        q.push(v);
      }
  }
  int res = s;
  for (int i = 1; i <= n; i++)
    if (d[i] > d[res]) res = i;
  return res;
}
void dfs(int u, int p) {
  par[u] = p;
  h[u] = h[p] + 1;
  for (auto v : a[u])
    if (v != par[u]) dfs(v, u);
}
void sfd(int u, int p, int cri, int curd) {
  for (auto v : a[u])
    if (v != p) {
      if (inpath[v])
        sfd(v, u, v, 0);
      else
        sfd(v, u, cri, curd + 1);
    }
  if (!inpath[u]) {
    res += max(d[cri], maxlen - d[cri]) + curd;
    if (d[cri] < maxlen - d[cri])
      ans[++cnt] = {u, t, u};
    else
      ans[++cnt] = {u, s, u};
  }
}
void solve() {
  s = bfs(1);
  t = bfs(s);
  dfs(1, 0);
  int ss = s, tt = t;
  if (h[ss] < h[tt]) swap(ss, tt);
  int hs = h[ss];
  for (int i = h[tt] + 1; i <= hs; i++) {
    inpath[ss] = 1;
    ss = par[ss];
  }
  while (ss != tt) {
    inpath[ss] = inpath[tt] = 1;
    ss = par[ss];
    tt = par[tt];
  }
  inpath[ss] = 1;
  maxlen = d[t];
  res = 1LL * d[t] * (d[t] + 1) / 2;
  sfd(ss, 0, ss, 0);
  tt = t;
  while (tt != s) {
    ans[++cnt] = {tt, s, tt};
    tt = trace[tt];
  }
  cout << res << '\n';
  for (int i = 1; i <= n - 1; i++)
    cout << ans[i].a << " " << ans[i].b << " " << ans[i].c << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  input();
  solve();
  return 0;
}