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 N = 2e5 + 5;
vector<pair<pair<int, int>, int> > vi;
vector<vector<int> > adj;
int h[N], f[N];
long long a[2][N], ans = 0;
bool vis[N];
void dfs(int u, int p) {
  f[u] = u;
  h[u] = 0;
  for (auto v : adj[u]) {
    if (v == p) continue;
    dfs(v, u);
    if (h[u] < h[v] + 1) {
      h[u] = h[v] + 1;
      f[u] = f[v];
    }
  }
}
void dfs1(int u, int p, int id) {
  for (auto v : adj[u]) {
    if (v == p) continue;
    a[id][v] = a[id][u] + 1;
    dfs1(v, u, id);
  }
}
void dfs2(int u, int p, int en) {
  vis[u] = false;
  if (u == en) {
    vis[u] = true;
    return;
  }
  for (auto v : adj[u]) {
    if (v == p) continue;
    dfs2(v, u, en);
    if (vis[v]) vis[u] = true;
  }
}
void dfs3(int u, int p, int st, int en) {
  for (auto v : adj[u]) {
    if (v == p) continue;
    dfs3(v, u, st, en);
  }
  if (!vis[u]) {
    if (a[0][u] > a[1][u]) {
      vi.push_back({{st, u}, u});
      ans += a[0][u];
    } else {
      vi.push_back({{en, u}, u});
      ans += a[1][u];
    }
  }
}
void dfs4(int u, int p, int st) {
  for (auto v : adj[u]) {
    if (v == p) continue;
    if (vis[v]) dfs4(v, u, st);
  }
  if (u != st) {
    vi.push_back({{st, u}, u});
  }
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  adj.assign(n + 1, vector<int>());
  for (int i = 1; i < n; i++) {
    int u, v;
    cin >> u >> v;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  dfs(1, -1);
  int st = f[1];
  memset(h, 0, sizeof(h));
  memset(f, 0, sizeof(f));
  dfs(st, -1);
  int en = f[st];
  memset(a, 0, sizeof(a));
  dfs1(st, -1, 0);
  dfs1(en, -1, 1);
  dfs2(st, -1, en);
  dfs3(st, -1, st, en);
  dfs4(st, -1, st);
  long long k = a[0][en];
  ans += k * (k + 1) / 2;
  cout << ans << "\n";
  for (auto tp : vi) {
    cout << tp.first.first << " " << tp.first.second << " " << tp.second
         << "\n";
  }
}