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 = 200100;
vector<long long> grafo[MAXN];
long long pai[MAXN];
pair<long long, long long> dfs(long long x, long long p, long long n) {
  pair<long long, long long> r = {x, n};
  for (long long i = 0; i < grafo[x].size(); i++) {
    long long at = grafo[x][i];
    if (at == p) continue;
    pai[at] = x;
    pair<long long, long long> h = dfs(at, x, n + 1);
    if (h.second > r.second) r = h;
  }
  return r;
}
long long d[5][MAXN];
void solve(long long x, long long p, long long n, long long k) {
  d[k][x] = n;
  for (long long i = 0; i < grafo[x].size(); i++) {
    long long at = grafo[x][i];
    if (at == p) continue;
    solve(at, x, n + 1, k);
  }
}
bool mark[MAXN];
void rec(long long x) {
  mark[x] = true;
  if (x == pai[x]) return;
  rec(pai[x]);
}
long long a, b;
void printa(long long x) {
  if (x == pai[x]) return;
  cout << a << " " << x << " " << x << endl;
  printa(pai[x]);
}
void get(long long x, long long p) {
  for (long long i = 0; i < grafo[x].size(); i++) {
    long long at = grafo[x][i];
    if (at == p || mark[at] == true) continue;
    get(at, x);
  }
  if (mark[x] == true) return;
  if (d[1][x] >= d[2][x])
    cout << a << " " << x << " " << x << endl;
  else
    cout << b << " " << x << " " << x << endl;
}
signed main() {
  long long n;
  cin >> n;
  for (long long i = 1; i < n; i++) {
    long long x, y;
    cin >> x >> y;
    grafo[x].push_back(y), grafo[y].push_back(x);
  }
  a = dfs(1, 1, 0).first;
  b = dfs(a, a, 0).first;
  pai[a] = a;
  rec(b);
  solve(a, a, 0, 1), solve(b, b, 0, 2);
  long long resp = 0;
  ;
  for (long long i = 1; i <= n; i++) {
    if (mark[i] == true) continue;
    resp += max(d[1][i], d[2][i]);
  }
  resp += (d[2][a] * (d[2][a] + 1)) / 2;
  cout << resp << endl;
  for (long long i = 1; i <= n; i++) {
    if (mark[i] == false) continue;
    get(i, i);
  }
  printa(b);
}