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;
namespace Acc {
const long long N = 2e5 + 10;
basic_string<long long> G[N];
long long n, st, bg, l, ed, f[N], dis[N], ans, pos, tag[N], stk[2][N], cnt;
void dfs(long long u, long long fa, long long d) {
  if (dis[u] = st ? d : 0, f[u] = fa, d > l) l = d, pos = u;
  for (auto v : G[u])
    if (v != fa) dfs(v, u, d + 1);
}
void print(long long u, long long fa, long long d) {
  for (auto v : G[u])
    if (v != fa) print(v, u, d + 1);
  if (!tag[u])
    ans += max(dis[u], d), cnt++, stk[0][cnt] = u,
                                  stk[1][cnt] = dis[u] > d ? bg : ed;
  else
    ans += d;
}
void work() {
  cin >> n;
  for (long long i = 1, x, y; i < n; i++) cin >> x >> y, G[x] += y, G[y] += x;
  dfs(1, 0, 0), bg = pos, st = 1, l = 0, dfs(bg, 0, 0), ed = pos, tag[bg] = 1;
  for (long long u = ed; u != bg; u = f[u]) tag[u] = 1;
  print(ed, 0, 0), cout << ans << endl;
  for (long long i = 1; i <= cnt; i++)
    cout << stk[0][i] << ' ' << stk[1][i] << ' ' << stk[0][i] << '\n';
  for (long long u = ed; u != bg; u = f[u])
    cout << u << ' ' << bg << ' ' << u << '\n';
}
}  // namespace Acc
int main() { return Acc::work(), 0; }