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>
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
const long long MOD = 1000000007;
const int INF = 1e9;
const long long LINF = 1LL * INF * INF;
const int MAXN = 200007;
const double EPS = 1e-7;
vector<int> G[MAXN];
bool used[MAXN], Del[MAXN];
int D1[MAXN], D2[MAXN], v1, v2;
long long s = 0;
vector<pair<int, int> > Ans;
void dfs1(int v, int *D) {
  used[v] = true;
  for (int i = (0); i < ((int)((G[v]).size())); ++i) {
    int to = G[v][i];
    if (!used[to]) {
      D[to] = D[v] + 1;
      dfs1(to, D);
    }
  }
}
bool dfs2(int v, bool d) {
  used[v] = true;
  Del[v] = v != v1 && (d || v != v2);
  for (int i = (0); i < ((int)((G[v]).size())); ++i) {
    int to = G[v][i];
    if (!used[to] && !Del[to] && !dfs2(to, d)) Del[v] = false;
  }
  if (Del[v]) {
    pair<int, int> p;
    if (D1[v] > D2[v] || d) {
      p.first = v1;
      s += D1[v];
    } else {
      p.first = v2;
      s += D2[v];
    }
    p.second = v;
    Ans.push_back(p);
  }
  return Del[v];
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  int n;
  cin >> n;
  for (int i = (0); i < (n - 1); ++i) {
    int a, b;
    cin >> a >> b;
    --a;
    --b;
    G[a].push_back(b);
    G[b].push_back(a);
  }
  D2[0] = 0;
  dfs1(0, D2);
  for (int i = (0); i < (n); ++i)
    if (D2[i] > D2[v1]) v1 = i;
  memset(used, 0, MAXN);
  D1[v1] = 0;
  dfs1(v1, D1);
  for (int i = (0); i < (n); ++i)
    if (D1[i] > D1[v2]) v2 = i;
  memset(used, 0, MAXN);
  D2[v2] = 0;
  dfs1(v2, D2);
  memset(used, 0, MAXN);
  dfs2(v1, 0);
  memset(used, 0, MAXN);
  dfs2(v1, 1);
  cout << s << '\n';
  for (int i = (0); i < (n - 1); ++i)
    cout << Ans[i].first + 1 << ' ' << Ans[i].second + 1 << ' '
         << Ans[i].second + 1 << '\n';
  return 0;
}