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;
vector<vector<int> > g;
vector<char> used;
vector<long long> distArr, fArr, sArr;
long long A, B, diameter;
long long answer = 0;
struct step {
  int a, b, d;
  friend ostream& operator<<(ostream& s, const step& v);
};
ostream& operator<<(ostream& s, const step& v) {
  s << v.a << " " << v.b << " " << v.d;
  return s;
}
vector<step> arr;
void dfs(int v, long long dist, vector<long long>& dArr, long long& cur) {
  dArr[v] = dist;
  if (dArr[cur] < dist) cur = v;
  for (int i = 0; i < (int)g[v].size(); i++) {
    int to = g[v][i];
    if (!used[to]) {
      used[to] = true;
      dfs(to, dist + 1, dArr, cur);
    }
  }
}
void dfs2(int v, long long dist, bool flag) {
  if (!flag) {
    sArr[v] = dist;
  }
  for (int i = 0; i < (int)g[v].size(); i++) {
    int to = g[v][i];
    if (!used[to]) {
      used[to] = true;
      dfs2(to, dist + 1, flag);
    }
  }
  if (fArr[v] + sArr[v] > diameter) {
    step temp;
    if (fArr[v] > sArr[v]) {
      temp.a = A + 1;
      answer += fArr[v];
    } else {
      temp.a = B + 1;
      answer += sArr[v];
    }
    temp.b = v + 1;
    temp.d = v + 1;
    arr.push_back(temp);
  }
  if (fArr[v] + sArr[v] == diameter && flag && v != A) {
    step temp;
    temp.a = A + 1;
    temp.d = v + 1;
    temp.b = v + 1;
    answer += fArr[v];
    arr.push_back(temp);
  } else if (fArr[v] + sArr[v] == diameter)
    used[v] = false;
}
int main() {
  int n;
  cin >> n;
  g.assign(n, vector<int>());
  for (int i = 0; i < n - 1; i++) {
    int from, to;
    cin >> from >> to;
    g[from - 1].push_back(to - 1);
    g[to - 1].push_back(from - 1);
  }
  used.assign(n, false);
  used[0] = true;
  distArr.assign(n, 0);
  A = 0;
  B = 0;
  dfs(0, 0, distArr, A);
  used.assign(n, false);
  used[A] = true;
  fArr.assign(n, 0);
  sArr.assign(n, 0);
  dfs(A, 0, fArr, B);
  diameter = fArr[B];
  used.assign(n, false);
  used[B] = true;
  dfs2(B, 0, false);
  dfs2(A, 0, true);
  cout << answer << endl;
  for (int i = 0; i < (int)arr.size(); i++) cout << arr[i] << endl;
  return 0;
}