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;
long long n;
vector<long long> e[200005];
long long dep[200005], p[200005], vis[200005];
vector<pair<long long, long long> > v;
long long mx, id, ans;
inline void addEdge(long long u, long long v) { e[u].push_back(v); }
void dfs(long long u, long long fa) {
  dep[u] = dep[fa] + 1, p[u] = fa;
  if (dep[u] > mx) mx = dep[u], id = u;
  for (long long v : e[u])
    if (v != fa) dfs(v, u);
}
void dfss(long long u, long long dep, long long fl, long long x) {
  vis[u] = 1;
  for (long long v : e[u])
    if (!vis[v]) dfss(v, dep + 1, 1, x);
  if (fl) ans += dep, v.push_back(make_pair(x, u));
}
inline long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (ch > '9' || ch < '0') {
    if (ch == '-') f = -1;
    ch = getchar();
    if (ch == -1) return 0;
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
signed main() {
  n = read();
  long long p1, p2;
  for (long long i = 1, u, v; i < n; i++)
    u = read(), v = read(), addEdge(u, v), addEdge(v, u);
  dfs(1, 0), mx = 0, p1 = id, dfs(id, 0), p2 = id, ans = mx * (mx - 1) / 2;
  for (long long i = p2; i; i = p[i]) vis[i] = 1;
  for (long long i = p2; i; i = p[i])
    if (dep[i] * 2 > dep[p2] + 1)
      dfss(i, dep[i] - 1, 0, p1);
    else
      dfss(i, dep[p2] - dep[i], 0, p2);
  cout << ans << endl;
  for (pair<long long, long long> it : v)
    printf("%lld %lld %lld\n", it.first, it.second, it.second);
  for (long long i = p2; i != p1; i = p[i])
    printf("%lld %lld %lld\n", p1, i, i);
  return 0;
}