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;
vector<int> graph[200005];
int dist[200005][2];
queue<int> qu;
int dgr[200005];
vector<pair<int, int>> v;
void dfs(int n, int p, int t) {
  for (int e : graph[n]) {
    if (e != p) {
      dist[e][t] = dist[n][t] + 1;
      dfs(e, n, t);
    }
  }
}
int getidx(int t) {
  int idx = 0;
  for (int i = 1; i <= N; i++) {
    idx = (dist[idx][t] < dist[i][t] ? i : idx);
  }
  return idx;
}
int main() {
  cin.sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> N;
  for (int i = 1; i < N; i++) {
    int a, b;
    cin >> a >> b;
    graph[a].push_back(b);
    graph[b].push_back(a);
  }
  dfs(1, 0, 0);
  int d1 = getidx(0);
  dist[d1][0] = 0;
  dfs(d1, 0, 0);
  int d2 = getidx(0);
  dfs(d2, 0, 1);
  for (int n = 1; n <= N; n++) {
    dgr[n] = graph[n].size();
    if (graph[n].size() == 1 && n != d1 && n != d2) {
      qu.push(n);
      dgr[n]--;
    }
  }
  long long ans = 0;
  while (qu.size()) {
    int n = qu.front();
    qu.pop();
    if (dist[n][0] > dist[n][1]) {
      ans += dist[n][0];
      v.push_back({d1, n});
    } else {
      ans += dist[n][1];
      v.push_back({d2, n});
    }
    for (int e : graph[n]) {
      if (dgr[e]) {
        dgr[e]--;
        if (dgr[e] == 1) {
          qu.push(e);
        }
      }
    }
  }
  ans += 1LL * dist[d1][1] * (dist[d1][1] + 1) / 2;
  qu.push(d1);
  while (qu.size()) {
    int n = qu.front();
    qu.pop();
    v.push_back({d2, n});
    for (int e : graph[n]) {
      if (dgr[e]) {
        dgr[e]--;
        if (dgr[e] == 1) {
          qu.push(e);
        }
      }
    }
  }
  cout << ans << "\n";
  for (int i = 0; i < N - 1; i++) {
    cout << v[i].first << " " << v[i].second << " " << v[i].second << "\n";
  }
}