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;
pair<int, int> dfs(const vector<vector<int>>& G, int v, int p) {
  pair<int, int> ret(0, v);
  for (auto v_ : G[v]) {
    if (v_ == p) continue;
    auto p = dfs(G, v_, v);
    ++p.first;
    ret = max(ret, p);
  }
  return ret;
}
void rec(const vector<vector<int>>& G, int v, int p, vector<int>& D) {
  for (auto v_ : G[v]) {
    if (v_ == p) continue;
    D[v_] = D[v] + 1;
    rec(G, v_, v, D);
  }
}
int main() {
  int n;
  scanf("%d", &n);
  vector<vector<int>> T(n);
  for (int i = 1; i < n; ++i) {
    int u, v;
    scanf("%d %d", &u, &v);
    --u, --v;
    T[u].push_back(v);
    T[v].push_back(u);
  }
  int l = dfs(T, 0, -1).second;
  int r = dfs(T, l, -1).second;
  vector<int> dl(n), dr(n);
  rec(T, l, -1, dl);
  rec(T, r, -1, dr);
  vector<pair<int, int>> W;
  vector<pair<int, int>> Q;
  long long ans = 0;
  for (int i = 0; i < n; ++i) {
    if (i == l) continue;
    if (dl[i] + dr[i] == dl[r]) {
      ans += dl[i];
      W.emplace_back(dl[i], i);
      continue;
    }
    ans += max(dl[i], dr[i]);
    Q.emplace_back(max(dl[i], dr[i]), i);
  }
  sort(Q.rbegin(), Q.rend());
  sort(W.rbegin(), W.rend());
  printf("%lld\n", ans);
  for (auto [d, i] : Q) {
    if (dl[i] > dr[i]) {
      printf("%d %d %d\n", l + 1, i + 1, i + 1);
    } else {
      printf("%d %d %d\n", r + 1, i + 1, i + 1);
    }
  }
  for (auto [d, i] : W) {
    printf("%d %d %d\n", l + 1, i + 1, i + 1);
  }
}