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;
int p[200010];
bool incoming[200010];
bool seen[200010];
bool inDiameter[200010];
int children[200010];
int d1[200010], d2[200010], d3[200010];
vector<vector<int> > edges;
int furthest[2];
int bestSoFar = 0;
void DFS(const int u, const int dist, const bool idx) {
  if (!idx)
    d3[u] = dist;
  else
    d1[u] = dist;
  if (dist > bestSoFar) furthest[idx] = u, bestSoFar = dist;
  seen[u] = true;
  for (int i = 0; i < (int)edges[u].size(); i++) {
    if (!seen[edges[u][i]]) {
      DFS(edges[u][i], dist + 1, idx);
      if (!idx) p[edges[u][i]] = u, children[u]++;
    }
  }
}
bool DFS2(const int u) {
  seen[u] = true;
  if (u == furthest[1]) {
    inDiameter[u] = true;
    return true;
  }
  for (int i = 0; i < (int)edges[u].size(); i++) {
    if (!seen[edges[u][i]] && DFS2(edges[u][i])) {
      inDiameter[u] = true;
      return true;
    }
  }
  return false;
}
void DFS3(const int u, const int dist) {
  d2[u] = dist;
  seen[u] = true;
  for (int i = 0; i < (int)edges[u].size(); i++)
    if (!seen[edges[u][i]]) DFS3(edges[u][i], dist + 1);
}
int main() {
  int N;
  int n1, n2;
  int root;
  queue<int> leafs;
  queue<pair<int, int> > chosen;
  int tot;
  long long res = 0;
  scanf("%d", &N);
  edges = vector<vector<int> >(N + 10);
  memset(p, 0, sizeof(p));
  memset(incoming, 0, sizeof(incoming));
  for (int i = 0; i < N - 1; i++) {
    scanf("%d %d", &n1, &n2);
    edges[n1].push_back(n2);
    incoming[n2] = true;
    edges[n2].push_back(n1);
  }
  for (int i = 1; i <= N; i++) {
    if (!incoming[i]) {
      root = i;
      break;
    }
  }
  memset(seen, 0, sizeof(seen));
  DFS(root, 0, 0);
  memset(seen, 0, sizeof(seen));
  bestSoFar = 0;
  DFS(furthest[0], 0, 1);
  memset(seen, 0, sizeof(seen));
  memset(inDiameter, 0, sizeof(inDiameter));
  DFS2(furthest[0]);
  memset(seen, 0, sizeof(seen));
  DFS3(furthest[1], 0);
  for (int i = 1; i <= N; i++)
    if (children[i] == 0 && !inDiameter[i]) leafs.push(i);
  while (!leafs.empty()) {
    tot = (int)leafs.size();
    while (tot--) {
      n1 = leafs.front();
      leafs.pop();
      n2 = d1[n1] > d2[n1] ? furthest[0] : furthest[1];
      chosen.push(pair<int, int>(n1, n2));
      res += max(d1[n1], d2[n1]);
      children[p[n1]]--;
      if (children[p[n1]] == 0 && !inDiameter[p[n1]]) leafs.push(p[n1]);
    }
  }
  int t = furthest[1];
  while (t != root && children[t] <= 1) {
    res += d1[t];
    chosen.push(pair<int, int>(t, furthest[0]));
    t = p[t];
  }
  t = furthest[0];
  while (p[t]) {
    res += d3[t];
    chosen.push(pair<int, int>(t, root));
    t = p[t];
  }
  printf("%I64d\n", res);
  while (!chosen.empty())
    printf("%d %d %d\n", chosen.front().first, chosen.front().second,
           chosen.front().first),
        chosen.pop();
  return 0;
}