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 int maxn = 2e5 + 10;
const int inf = 1e9 + 7;
int head[maxn], to[maxn << 1], nxt[maxn << 1];
int Vertex;
int qu[maxn], qf, qb, dist[2][maxn], fa[maxn], i, v, ok;
int dijk(int u, int *Dist) {
  for (i = 1; i <= Vertex; i++) Dist[i] = inf;
  Dist[u] = 0;
  qf = qb = 0;
  qu[qb++] = u;
  while (qf < qb) {
    u = qu[qf++];
    for (i = head[u]; i; i = nxt[i]) {
      v = to[i];
      if (Dist[v] > Dist[u] + 1) {
        if (ok) fa[v] = u;
        Dist[v] = Dist[u] + 1;
        qu[qb++] = v;
      }
    }
  }
  return qu[qf - 1];
}
int u, U, V, E;
int vis[maxn];
long long answer;
vector<pair<pair<int, int>, int> > res;
int main() {
  scanf("%d", &Vertex);
  for (i = 1; i < Vertex; i++) {
    scanf("%d%d", &u, &v);
    nxt[++E] = head[u];
    to[E] = v;
    head[u] = E;
    nxt[++E] = head[v];
    to[E] = u;
    head[v] = E;
  }
  U = dijk(1, dist[0]);
  V = dijk(U, dist[0]);
  ok = 1;
  dijk(V, dist[1]);
  u = U;
  while (u) {
    vis[u] = 1;
    u = fa[u];
  }
  for (i = qf - 1; i >= 0; i--) {
    u = qu[i];
    if (!vis[u]) {
      if (dist[0][u] > dist[1][u]) {
        answer += dist[0][u];
        res.push_back(make_pair(make_pair(u, U), u));
      } else {
        answer += dist[1][u];
        res.push_back(make_pair(make_pair(u, V), u));
      }
    }
  }
  u = U;
  while (u != V) {
    answer += dist[1][u];
    res.push_back(make_pair(make_pair(u, V), u));
    u = fa[u];
  }
  cout << answer << endl;
  for (i = 0; i < ((int)((res).size())); i++) {
    printf("%d %d %d\n", res[i].first.first, res[i].first.second,
           res[i].second);
  }
  return 0;
}