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;
template <typename T, typename U>
inline void smin(T &a, U b) {
  if (a > b) a = b;
}
template <typename T, typename U>
inline void smax(T &a, U b) {
  if (a < b) a = b;
}
template <class T>
inline void gn(T &first) {
  char c, sg = 0;
  while (c = getchar(), (c > '9' || c < '0') && c != '-')
    ;
  for ((c == '-' ? sg = 1, c = getchar() : 0), first = 0; c >= '0' && c <= '9';
       c = getchar())
    first = (first << 1) + (first << 3) + c - '0';
  if (sg) first = -first;
}
template <class T1, class T2>
inline void gn(T1 &x1, T2 &x2) {
  gn(x1), gn(x2);
}
template <class T1, class T2, class T3>
inline void gn(T1 &x1, T2 &x2, T3 &x3) {
  gn(x1), gn(x2), gn(x3);
}
template <class T1, class T2, class T3, class T4>
inline void gn(T1 &x1, T2 &x2, T3 &x3, T4 &x4) {
  gn(x1), gn(x2), gn(x3), gn(x4);
}
template <class T>
inline void print(T first) {
  if (first < 0) {
    putchar('-');
    return print(-first);
  }
  if (first < 10) {
    putchar('0' + first);
    return;
  }
  print(first / 10);
  putchar(first % 10 + '0');
}
template <class T>
inline void printsp(T first) {
  print(first);
  putchar(' ');
}
template <class T>
inline void println(T first) {
  print(first);
  putchar('\n');
}
template <class T1, class T2>
inline void print(T1 x1, T2 x2) {
  printsp(x1), println(x2);
}
template <class T1, class T2, class T3>
inline void print(T1 x1, T2 x2, T3 x3) {
  printsp(x1), printsp(x2), println(x3);
}
template <class T1, class T2, class T3, class T4>
inline void print(T1 x1, T2 x2, T3 x3, T4 x4) {
  printsp(x1), printsp(x2), printsp(x3), println(x4);
}
int power(int a, int b, int m, int ans = 1) {
  for (; b; b >>= 1, a = 1LL * a * a % m)
    if (b & 1) ans = 1LL * ans * a % m;
  return ans;
}
pair<int, int> ans[200010];
vector<int> adj[200010];
queue<int> q;
int d[200010], d1[200010], d2[200010];
int pre[200010], vst[200010];
int nn, n, A, B;
long long Max;
int end_point(int id, int first) {
  memset(d, 0x3f3f3f3f, sizeof d);
  memset(pre, 0, sizeof pre);
  q.push(id);
  d[id] = 0;
  while (!q.empty()) {
    int u = q.front();
    q.pop();
    for (int i = 0; i < adj[u].size(); i++) {
      int v = adj[u][i];
      if (d[v] > d[u] + 1) {
        pre[v] = u;
        d[v] = d[u] + 1;
        q.push(v);
      }
    }
  }
  for (int i = 1; i <= n; i++) {
    if (first)
      d1[i] = d[i];
    else
      d2[i] = d[i];
  }
  int mx = *max_element(d + 1, d + n + 1);
  for (int i = 1; i <= n; i++)
    if (d[i] == mx) {
      return i;
    }
}
void dfs(int u, int pa = -1) {
  for (int i = 0; i < adj[u].size(); i++) {
    int v = adj[u][i];
    if (v != pa) {
      if (vst[v] == 0) {
        if (d1[v] > d2[v])
          ans[nn++] = pair<int, int>(v, A);
        else
          ans[nn++] = pair<int, int>(v, B);
        Max += max(d1[v], d2[v]);
      } else
        Max += d2[v];
      dfs(v, u);
    }
  }
}
int main() {
  gn(n);
  for (int i = 1; i < n; i++) {
    int u, v;
    gn(u, v);
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  A = end_point(1, 1);
  B = end_point(A, 1);
  A = end_point(B, 0);
  int u = A;
  while (u != B) {
    ans[nn++] = pair<int, int>(u, B);
    vst[u] = 1;
    u = pre[u];
  }
  reverse(ans, ans + nn);
  dfs(B);
  println(Max);
  for (int i = nn - 1; i >= 0; i--) {
    print(ans[i].first, ans[i].second, ans[i].first);
  }
}