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;
const long long MOD = 1e9 + 7;
const long long MAXN = 1e6 + 10;
const long long INF = 8e18;
const long long LOG = 19;
vector<long long> adj[MAXN], l;
bool mark[MAXN];
long long di[MAXN], d1[MAXN], d2[MAXN], d3[MAXN], par[MAXN], y, x, ans;
pair<long long, long long> dm;
vector<pair<long long, pair<long long, long long>>> res;
void dfs(long long u, long long p, long long k, long long d[]) {
  for (auto i : adj[u]) {
    if (i != p) {
      par[i] = u;
      d[i] = d[par[i]] + 1;
      dfs(i, u, k + 1, d);
    }
  }
}
void diam(long long u, long long p, long long k) {
  if (x < k) {
    x = k;
    y = u;
  }
  for (auto i : adj[u])
    if (p != i) diam(i, u, k + 1);
}
void haras(vector<long long> l) {
  long long u;
  while (!l.empty()) {
    u = l.back();
    di[u] = -1;
    l.pop_back();
    if (d1[u] < d2[u]) {
      ans += d2[u];
      res.push_back({u, {dm.second, u}});
    } else {
      ans += d1[u];
      res.push_back({u, {dm.first, u}});
    }
    for (auto i : adj[u]) {
      di[i]--;
      if (di[i] == 1) l.push_back(i);
    }
  }
}
void p_haras(long long u, long long v) {
  while (u != v) {
    ans += d1[v];
    res.push_back({v, {u, v}});
    v = par[v];
  }
}
void solve(long long n) {
  dfs(dm.second, -1, 0, d2);
  dfs(dm.first, -1, 0, d1);
  haras(l);
  p_haras(dm.first, dm.second);
}
int main() {
  ;
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  long long n;
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    long long u, v;
    cin >> u >> v;
    di[u]++;
    di[v]++;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  diam(1, -1, 0);
  dm.first = y;
  x = 0;
  y = 0;
  diam(dm.first, -1, 0);
  dm.second = y;
  y = 0;
  x = 0;
  for (int i = 1; i <= n; i++) {
    if (i == dm.first || i == dm.second) continue;
    if (di[i] == 1) {
      l.push_back(i);
    }
  }
  solve(n);
  cout << ans << '\n';
  for (auto i : res) {
    cout << i.first << " " << i.second.first << " " << i.second.second << '\n';
  }
  return 0;
}