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 maxn = 2e5 + 1;
vector<int> g[maxn];
int d1[maxn];
int d2[maxn];
int nei[maxn];
bool was[maxn];
int s, t;
void dfs1(int v, int p = 0, int dist = 0) {
  d1[v] = max(d1[v], dist);
  for (int x : g[v]) {
    if (x != p) dfs1(x, v, dist + 1);
  }
}
void dfs2(int v, int p = 0, int dist = 0) {
  d2[v] = max(d2[v], dist);
  for (int x : g[v]) {
    if (x != p) dfs2(x, v, dist + 1);
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  for (int i = 1; i < n; i++) {
    int a, b;
    cin >> a >> b;
    g[a].push_back(b);
    g[b].push_back(a);
    nei[a]++, nei[b]++;
  }
  dfs1(1);
  s = max_element(d1, d1 + n + 1) - d1;
  fill(d1, d1 + n + 1, 0);
  dfs1(s);
  t = max_element(d1, d1 + n + 1) - d1;
  dfs2(t);
  queue<pair<int, int>> q;
  for (int i = 1; i <= n; i++) {
    if (nei[i] == 1 and i != s and i != t) q.push({max(d1[i], d2[i]), i});
  }
  vector<pair<pair<int, int>, int>> a;
  long long ans = 0;
  while (!q.empty()) {
    auto x = q.front();
    q.pop();
    ans += x.first;
    nei[x.second]--;
    if (d1[x.second] > d2[x.second])
      a.push_back({{s, x.second}, x.second});
    else
      a.push_back({{t, x.second}, x.second});
    for (int v : g[x.second]) {
      nei[v]--;
      if (nei[v] == 1) q.push({max(d1[v], d2[v]), v});
    }
  }
  while (nei[t] != 0) {
    a.push_back({{s, t}, t});
    ans += d1[t];
    nei[t]--;
    for (int x : g[t]) {
      nei[x]--;
      if (nei[x] == 1) {
        t = x;
        break;
      }
    }
  }
  cout << ans << endl;
  for (auto x : a)
    cout << x.first.first << ' ' << x.first.second << ' ' << x.second << '\n';
  return 0;
}