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 MOD = 998244353;
const double oula = 0.57721566490153286060651209;
using namespace std;
int n;
vector<int> ve[200005];
vector<pair<int, pair<int, int> > > ans;
int deep[200005], fa[200005], book[200005];
long long sum;
int pos1, pos2;
void dfs1(int now, int pre) {
  for (int i = 0; i < ve[now].size(); i++) {
    if (ve[now][i] != pre) {
      deep[ve[now][i]] = deep[now] + 1;
      fa[ve[now][i]] = now;
      dfs1(ve[now][i], now);
    }
  }
}
void dfs2(int now, int pre, int d) {
  if (book[now]) d++;
  for (int i = 0; i < ve[now].size(); i++) {
    if (pre != ve[now][i]) {
      dfs2(ve[now][i], now, d);
      if (!book[ve[now][i]]) {
        int dep1 = deep[ve[now][i]];
        int dep2 = deep[pos2] + deep[ve[now][i]] - 2 * d;
        if (dep1 >= dep2) {
          sum += dep1;
          ans.push_back({pos1, {ve[now][i], ve[now][i]}});
        } else {
          sum += dep2;
          ans.push_back({pos2, {ve[now][i], ve[now][i]}});
        }
      }
    }
  }
}
int main() {
  std::ios::sync_with_stdio(false);
  cin >> n;
  int u, v;
  for (int i = 1; i < n; i++) {
    cin >> u >> v;
    ve[u].push_back(v);
    ve[v].push_back(u);
  }
  dfs1(1, 0);
  pos1 = 0, pos2 = 0;
  for (int i = 1; i <= n; i++) {
    if (deep[i] > deep[pos1]) {
      pos1 = i;
    }
  }
  deep[pos1] = 0, fa[pos1] = 0;
  dfs1(pos1, 0);
  for (int i = 1; i <= n; i++) {
    if (deep[i] > deep[pos2]) {
      pos2 = i;
    }
  }
  for (int i = pos2; i != pos1; i = fa[i]) book[i] = 1;
  dfs2(pos1, 0, 0);
  for (int i = pos2; i != pos1; i = fa[i]) {
    ans.push_back({pos1, {i, i}});
    sum += deep[i];
  }
  cout << sum << endl;
  for (int i = 0; i < ans.size(); i++) {
    cout << ans[i].first << " " << ans[i].second.first << " "
         << ans[i].second.second << endl;
  }
}