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 = 2 * 1e5 + 17;
long long R;
bool rem[maxn];
int N, H, V1, V2;
int deg[maxn], DPV1[maxn], DPV2[maxn];
vector<pair<int, int> > res;
vector<int> leaves, graph[maxn];
void readEdge() {
  int u, v;
  scanf("%d %d", &u, &v);
  deg[u]++, deg[v]++;
  graph[u].push_back(v), graph[v].push_back(u);
}
void DFS(int idx, int prev, int len, int &where, int *DP) {
  DP[idx] = len;
  if (DP[idx] > DP[where]) where = idx;
  for (int u : graph[idx])
    if (u != prev) DFS(u, idx, len + 1, where, DP);
}
int main() {
  scanf("%d", &N);
  for (int i = 1; i < N; i++) readEdge();
  DFS(1, 1, 0, V1, DPV2);
  DFS(V1, V1, 0, V2, DPV1);
  memset(DPV2, 0, sizeof(DPV2));
  DFS(V2, V2, 0, H, DPV2);
  for (int i = 1; i <= N; i++)
    if (i != V1 && i != V2 && deg[i] == 1) leaves.push_back(i);
  while (leaves.size()) {
    int idx = leaves.back();
    leaves.pop_back();
    rem[idx] = true;
    R += max(DPV1[idx], DPV2[idx]);
    res.push_back(make_pair((DPV1[idx] > DPV2[idx] ? V1 : V2), idx));
    for (int v : graph[idx])
      if (!rem[v] && --deg[v] == 1) leaves.push_back(v);
  }
  leaves.push_back(V2);
  while (leaves.size()) {
    int idx = leaves.back();
    leaves.pop_back();
    rem[idx] = true;
    R += DPV1[idx];
    res.push_back(make_pair(V1, idx));
    for (int v : graph[idx])
      if (!rem[v] && --deg[v] == 1) leaves.push_back(v);
  }
  printf("%I64d\n", R);
  for (auto p : res) printf("%d %d %d\n", p.first, p.second, p.second);
  return 0;
}