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 Z = (int)2e5 + 111;
const int inf = (int)1e9 + 111;
const long long llinf = (long long)1e18 + 5;
const int MOD = (int)1e9 + 7;
int timer, tin[Z], tout[Z], up[Z][44], h[Z], d[Z], l, ok, v1, v2, pr[Z];
vector<int> path;
set<int> g[Z];
set<pair<int, int> > mn;
void dfs(int v, int p, int depth) {
  pr[v] = p;
  tin[v] = ++timer;
  up[v][0] = p;
  h[v] = depth;
  for (int i = 1; i <= l; ++i) {
    up[v][i] = up[up[v][i - 1]][i - 1];
  }
  for (int to : g[v]) {
    if (to == p) continue;
    dfs(to, v, depth + 1);
  }
  tout[v] = ++timer;
}
bool is(int x, int y) { return tin[x] <= tin[y] && tout[x] >= tout[y]; }
int lca(int x, int y) {
  if (is(x, y)) return x;
  if (is(y, x)) return y;
  for (int i = l; i >= 0; --i) {
    if (!is(up[x][i], y)) {
      x = up[x][i];
    }
  }
  return up[x][0];
}
int dist(int x, int y) { return h[x] + h[y] - 2 * h[lca(x, y)]; }
void dfs2(int v, int p, int depth) {
  d[v] = depth;
  for (int to : g[v]) {
    if (to == p) continue;
    dfs2(to, v, depth + 1);
  }
}
void dfs3(int v, int p) {
  if (v == v2 || ok) {
    ok = 1;
    return;
  }
  for (int to : g[v]) {
    if (to == p) continue;
    dfs3(to, v);
  }
  if (ok) {
    path.push_back(v);
  }
}
int main() {
  srand(time(0));
  ios_base::sync_with_stdio(false);
  int n;
  cin >> n;
  while ((1 << l) <= n) ++l;
  for (int i = 1; i < n; ++i) {
    int x, y;
    cin >> x >> y;
    g[x].insert(y);
    g[y].insert(x);
  }
  dfs(1, 1, 0);
  v1 = 1;
  dfs2(v1, v1, 0);
  for (int i = 1; i <= n; ++i) {
    if (d[i] > d[v1]) {
      v1 = i;
    }
  }
  dfs2(v1, v1, 0);
  v2 = 1;
  for (int i = 1; i <= n; ++i) {
    if (d[i] > d[v2]) {
      v2 = i;
    }
  }
  dfs3(v1, v1);
  reverse((path).begin(), (path).end());
  path.push_back(v2);
  for (int i = 1; i <= n; ++i) {
    mn.insert(make_pair((int)g[i].size(), i));
  }
  for (int v : path) {
    mn.erase(make_pair((int)g[v].size(), v));
  }
  long long ans = 1LL * (int)path.size() * ((int)path.size() - 1) / 2;
  vector<pair<pair<int, int>, int> > vecAns;
  while (!mn.empty()) {
    int v = mn.begin()->second;
    if (dist(v1, v) > dist(v2, v)) {
      vecAns.push_back(make_pair(make_pair(v1, v), v));
      ans += dist(v1, v);
    } else {
      ans += dist(v, v2);
      vecAns.push_back(make_pair(make_pair(v2, v), v));
    }
    mn.erase(mn.begin());
    for (int to : g[v]) {
      if (mn.count(make_pair((int)g[to].size(), to))) {
        mn.erase(make_pair((int)g[to].size(), to));
        g[to].erase(v);
        mn.insert(make_pair((int)g[to].size(), to));
      }
    }
  }
  for (int i = (int)path.size() - 1; i > 0; --i) {
    vecAns.push_back(make_pair(make_pair(path[0], path[i]), path[i]));
  }
  cout << ans << '\n';
  for (auto i : vecAns) {
    cout << i.first.first << ' ' << i.first.second << ' ' << i.second << '\n';
  }
  return 0;
}