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;
using ll = long long;
const int MX = 2e5 + 1;
vector<int> adj[MX], dist1, dist2;
vector<array<int, 2>> ops;
bool save[MX];
int par[MX], best;
int pt1, pt2;
ll ans;
void dfs(int u, vector<int>& dep) {
  for (auto c : adj[u]) {
    if (c != par[u]) {
      dep[c] = dep[u] + 1;
      par[c] = u;
      dfs(c, dep);
    }
  }
}
void dfs_ans(int u) {
  for (auto c : adj[u]) {
    if (c != par[u]) {
      dfs_ans(c);
    }
  }
  if (!save[u]) {
    if (dist1[u] > dist2[u]) {
      ops.push_back({pt1, u});
      ans += dist1[u];
    } else {
      ops.push_back({pt2, u});
      ans += dist2[u];
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  for (int i = 0; i < n - 1; ++i) {
    int u, v;
    cin >> u >> v;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  dist1.resize(n + 1);
  dfs(1, dist1);
  pt1 = max_element(dist1.begin(), dist1.end()) - dist1.begin();
  dist1[pt1] = 0;
  par[pt1] = 0;
  dfs(pt1, dist1);
  pt2 = max_element(dist1.begin(), dist1.end()) - dist1.begin();
  dist2.resize(n + 1);
  par[pt2] = 0;
  dfs(pt2, dist2);
  for (int i = pt1; i; i = par[i]) {
    save[i] = true;
  }
  dfs_ans(pt2);
  for (int i = pt1; i != pt2; i = par[i]) {
    ops.push_back({pt2, i});
    ans += dist2[i];
  }
  cout << ans << '\n';
  for (auto i : ops) {
    cout << i[0] << ' ' << i[1] << ' ' << i[1] << '\n';
  }
}