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 long long N = 2e5 + 5;
vector<long long> G[N];
long long p[N], vis[N], d[N], m[N], cnt = 0, D[N], x, y;
vector<vector<long long>> ans;
void dfs(long long v, long long pa) {
  vis[v] = 1;
  p[v] = pa;
  d[v] = d[pa] + 1;
  for (long long i : G[v]) {
    if (!vis[i] && i != pa) {
      p[i] = v;
      dfs(i, v);
    }
  }
}
void dfs2(long long v, long long pa) {
  vis[v] = 1;
  if (m[v]) {
    D[v] = d[v];
  } else
    D[v] = D[pa];
  for (long long i : G[v])
    if (!vis[i] && i != pa) dfs2(i, v);
}
void dfs3(long long v, long long pa) {
  vis[v] = 1;
  for (long long i : G[v])
    if (!vis[i] && i != pa) dfs3(i, v);
  if (!m[v]) {
    long long pd = d[v] - D[v], par;
    if (D[v] - 1 > D[y] - D[v]) {
      par = x;
      pd += D[v] - 1;
    } else {
      par = y;
      pd += D[y] - D[v];
    }
    ans.push_back({v, par, v});
    cnt += pd;
  }
}
int main() {
  memset(m, 0, sizeof(m));
  memset(d, 0, sizeof(d));
  memset(vis, 0, sizeof(vis));
  memset(p, 0, sizeof(p));
  memset(D, 0, sizeof(D));
  long long n;
  cin >> n;
  for (int i = 1; i < n; i++) {
    long long u, v;
    cin >> u >> v;
    G[u].push_back(v);
    G[v].push_back(u);
  }
  p[1] = 1;
  dfs(1, 1);
  long long mx = 0;
  x = 1;
  for (int i = 1; i <= n; i++) mx = max(mx, d[i]);
  for (int i = 1; i <= n; i++)
    if (d[i] == mx) x = i;
  memset(d, 0, sizeof(d));
  memset(vis, 0, sizeof(vis));
  memset(p, 0, sizeof(p));
  p[x] = x;
  dfs(x, x);
  mx = 0;
  for (int i = 1; i <= n; i++) mx = max(mx, d[i]);
  for (int i = 1; i <= n; i++)
    if (d[i] == mx) y = i;
  long long j = y, len = 0;
  while (j != x) {
    m[j] = 1;
    j = p[j];
    len++;
  }
  m[x] = 1;
  len++;
  memset(vis, 0, sizeof(vis));
  dfs2(x, x);
  memset(vis, 0, sizeof(vis));
  dfs3(x, x);
  j = y;
  while (j != x) {
    ans.push_back({j, x, j});
    j = p[j];
  }
  cnt += (len * (len - 1)) / 2;
  cout << cnt << endl;
  for (vector<long long> V : ans) {
    for (long long i : V) cout << i << " ";
    cout << endl;
  }
}