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 n, a, b, dout[200200];
vector<int> graph[200200];
int aux, atuald;
void dfs(int u, int p, int d) {
  if (d > atuald) {
    aux = u;
    atuald = d;
  }
  for (int v : graph[u])
    if (v != p) dfs(v, u, d + 1);
}
int t, dist[2][200200];
void dfs1(int u, int p, int d) {
  dist[t][u] = d;
  for (int v : graph[u])
    if (v != p) dfs1(v, u, d + 1);
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n - 1; i++) {
    scanf("%d %d", &a, &b);
    graph[a].push_back(b);
    graph[b].push_back(a);
    dout[a]++;
    dout[b]++;
  }
  pair<int, int> diam;
  aux = 1, atuald = 0;
  dfs(1, 0, 0);
  diam.first = aux;
  atuald = 0;
  dfs(aux, 0, 0);
  diam.second = aux;
  for (int i = 0; i <= n; i++) dist[0][i] = dist[1][i] = 100000000;
  t = 0;
  dfs1(diam.first, 0, 0);
  t = 1;
  dfs1(diam.second, 0, 0);
  long long ans = 0;
  vector<pair<int, int> > chosen;
  deque<int> leafs;
  for (int i = 1; i <= n; i++)
    if (dout[i] == 1) leafs.push_back(i);
  while (leafs.size() != 2) {
    while (leafs.back() == diam.first || leafs.back() == diam.second) {
      leafs.push_front(leafs.back());
      leafs.pop_back();
    }
    int x = leafs.back();
    leafs.pop_back();
    if (dist[0][x] > dist[1][x]) {
      chosen.push_back(pair<int, int>(diam.first, x));
      ans += dist[0][x];
    } else {
      chosen.push_back(pair<int, int>(diam.second, x));
      ans += dist[1][x];
    }
    for (int v : graph[x]) {
      dout[v]--;
      if (dout[v] == 1) leafs.push_back(v);
    }
  }
  if (leafs[0] == diam.second) swap(leafs[0], leafs[1]);
  while (leafs.size() != 1) {
    int x = leafs[1];
    leafs.pop_back();
    chosen.push_back(pair<int, int>(leafs[0], leafs[1]));
    ans += dist[0][leafs[1]];
    for (int v : graph[x]) {
      dout[v]--;
      if (dout[v] == 1) leafs.push_back(v);
    }
  }
  printf("%lld\n", ans);
  for (int i = 0; i < chosen.size(); i++)
    printf("%d %d %d\n", chosen[i].first, chosen[i].second, chosen[i].second);
  return 0;
}