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>
const int N = 500050;
using namespace std;
inline int gi() {
  char ch = getchar();
  int x = 0, q = 0;
  while (ch < '0' || ch > '9') ch == '-' ? q = 1 : 0, ch = getchar();
  while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
  return q ? (-x) : x;
}
int S, T, u, v, nn[N][2], st[N], head[N], dis[N], fa[N], bj[N], cnt, top, tot;
void bfs() {
  memset(dis, 0, sizeof(dis));
  int l = 0, r = 1, s, t, mx = 0;
  dis[st[1] = S] = 1;
  while (l < r) {
    s = st[++l];
    if (dis[s] > mx) T = s, mx = dis[s];
    for (int i = head[s]; i; i = nn[i][0])
      if (!dis[t = nn[i][1]]) dis[t] = dis[s] + 1, st[++r] = t, fa[t] = s;
  }
  return;
}
void dfs(int x, int fu, int d) {
  if (d > dis[x]) dis[x] = d, fa[x] = S;
  for (int i = head[x], y; i; i = nn[i][0])
    if ((y = nn[i][1]) != fu) dfs(y, x, d + 1);
  return;
}
void cal(int x, int fu) {
  for (int i = head[x], y; i; i = nn[i][0])
    if ((y = nn[i][1]) != fu) cal(y, x);
  if (!bj[x]) printf("%d %d %d\n", x, fa[x], x);
  return;
}
int main() {
  int n = gi(), u, v;
  long long ans = 0;
  for (int i = 2; i <= n; ++i)
    u = gi(), v = gi(), (nn[++cnt][1] = v, nn[cnt][0] = head[u], head[u] = cnt),
    (nn[++cnt][1] = u, nn[cnt][0] = head[v], head[v] = cnt);
  S = 1, bfs();
  S = T, bfs();
  ans = 1ll * dis[T] * (dis[T] - 1) / 2;
  fa[S] = 0;
  for (int i = T; i; i = fa[i]) bj[st[++tot] = i] = 1;
  memset(dis, 0, sizeof(dis));
  dfs(S, 0, 0), S = T, dfs(T, 0, 0);
  for (int i = 1; i <= n; ++i)
    if (!bj[i]) ans += dis[i];
  cout << ans << endl;
  cal(S, 0);
  for (int i = 1; i < tot; ++i) printf("%d %d %d\n", st[i], st[tot], st[i]);
  return 0;
}