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>
inline void read(T& s) {
  s = 0;
  int f = 0;
  char c = getchar();
  while (!isdigit(c)) f |= (c == '-'), c = getchar();
  while (isdigit(c)) s = s * 10 + (c ^ 48), c = getchar();
  s = f ? -s : s;
}
const int _ = 200010;
const int __ = 400010;
int tot, head[_], nxt[__], ver[__];
inline void Add_edge(int u, int v) {
  nxt[++tot] = head[u], head[u] = tot, ver[tot] = v;
}
int n, rt, lf, mark[_], fa[_], dep[_];
queue<pair<int, int> > Q;
inline void dfs1(int u, int f) {
  for (register int i = head[u]; i; i = nxt[i]) {
    int v = ver[i];
    if (v == f) continue;
    dep[v] = dep[u] + 1, dfs1(v, u);
  }
}
inline void dfs2(int u, int f) {
  for (register int i = head[u]; i; i = nxt[i]) {
    int v = ver[i];
    if (v == f) continue;
    fa[v] = u, dfs2(v, u), mark[u] |= mark[v];
  }
}
long long ans = 0;
inline void dfs3(int u, int f, int lca) {
  for (register int i = head[u]; i; i = nxt[i]) {
    int v = ver[i];
    if (v == f) continue;
    dfs3(v, u, mark[v] ? v : lca);
  }
  if (!mark[u]) {
    if (dep[u] > dep[u] + dep[lf] - 2 * dep[lca])
      ans += dep[u], Q.push(make_pair(rt, u));
    else
      ans += dep[u] + dep[lf] - 2 * dep[lca], Q.push(make_pair(lf, u));
  }
}
int main() {
  read(n);
  for (register int u, v, i = 1; i < n; ++i)
    read(u), read(v), Add_edge(u, v), Add_edge(v, u);
  dep[1] = 0, dfs1(1, 0);
  rt = 1;
  for (register int i = 1; i <= n; ++i)
    if (dep[i] > dep[rt]) rt = i;
  dep[rt] = 0, dfs1(rt, 0);
  lf = rt;
  for (register int i = 1; i <= n; ++i)
    if (dep[i] > dep[lf]) lf = i;
  mark[lf] = 1;
  dfs2(rt, 0);
  dfs3(rt, 0, rt);
  for (register int i = lf; i != rt; i = fa[i])
    ans += dep[i], Q.push(make_pair(rt, i));
  printf("%lld\n", ans);
  while (!Q.empty()) {
    pair<int, int> x = Q.front();
    Q.pop();
    printf("%d %d %d\n", x.first, x.second, x.second);
  }
  return 0;
}