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 N = 2e5 + 5;
vector<int> adj[N];
int degree[N], dist_from_1st_node[N], dist_from_2nd_node[N];
int main(void) {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n;
  cin >> n;
  int m = n - 1;
  while (m--) {
    int u, v;
    cin >> u >> v;
    adj[u].push_back(v);
    adj[v].push_back(u);
    ++degree[u];
    ++degree[v];
  }
  function<void(int, int, int*)> dfs;
  dfs = [&](int u, int parent, int* dist) -> void {
    for (int v : adj[u]) {
      if (v == parent) {
        continue;
      }
      dist[v] = dist[u] + 1;
      dfs(v, u, dist);
    }
  };
  dfs(1, 0, dist_from_1st_node);
  int x2 = max_element(dist_from_1st_node + 1, dist_from_1st_node + n + 1) -
           dist_from_1st_node;
  dfs(x2, 0, dist_from_2nd_node);
  int x1 = max_element(dist_from_2nd_node + 1, dist_from_2nd_node + n + 1) -
           dist_from_2nd_node;
  dist_from_1st_node[x1] = 0;
  dfs(x1, 0, dist_from_1st_node);
  queue<int> q;
  for (int i = 1; i <= n; ++i) {
    if (i != x1 && i != x2 && degree[i] == 1) {
      q.push(i);
    }
  }
  vector<pair<int, int>> ans;
  long long sum = 0;
  while (!q.empty()) {
    int victim = q.front();
    q.pop();
    if (dist_from_1st_node[victim] > dist_from_2nd_node[victim]) {
      ans.push_back(make_pair(x1, victim));
      sum += dist_from_1st_node[victim];
    } else {
      ans.push_back(make_pair(x2, victim));
      sum += dist_from_2nd_node[victim];
    }
    for (int neighbor : adj[victim]) {
      --degree[neighbor];
      if (degree[neighbor] == 1) {
        q.push(neighbor);
      }
    }
  }
  int x = x1;
  while (x != x2) {
    sum += dist_from_2nd_node[x];
    ans.push_back(make_pair(x2, x));
    for (int neighbor : adj[x]) {
      --degree[neighbor];
      if (degree[neighbor] == 1 || neighbor == x2) {
        x = neighbor;
        break;
      }
    }
  }
  cout << sum << "\n";
  for (pair<int, int>& pair : ans) {
    cout << pair.first << " " << pair.second << " " << pair.second << "\n";
  }
  return 0;
}