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 INF = (int)2e9 + 123;
const long long LINF = (long long)1e18 + 123;
const long double EPS = (long double)1e-7;
const long long MOD = (long long)1e9 + 7;
long long power(long long x, long long y, long long mod = MOD) {
  if (y == 0) {
    return 1;
  }
  if (y & 1) {
    return power(x, y - 1, mod) * x % mod;
  } else {
    long long tmp = power(x, y / 2, mod);
    return tmp * tmp % mod;
  }
}
template <typename A, typename B>
bool mini(A &x, const B &y) {
  if (y < x) {
    x = y;
    return true;
  }
  return false;
}
template <typename A, typename B>
bool maxi(A &x, const B &y) {
  if (y > x) {
    x = y;
    return true;
  }
  return false;
}
void add(long long &x, long long y) {
  x += y;
  if (x >= MOD) x -= MOD;
  if (x < 0) x += MOD;
}
long long mult(long long x, long long y) { return x * y % MOD; }
void run();
int main() {
  if (strlen("")) {
    freopen(
        ""
        ".in",
        "r", stdin);
    freopen(
        ""
        ".out",
        "w", stdout);
  }
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cout << fixed << setprecision(12);
  run();
  return 0;
}
const int N = (int)2e5 + 123;
int n;
vector<int> g[N];
bool in_diam[N];
pair<int, int> farest(int v, int from = -1) {
  pair<int, int> res = {0, v};
  for (int u : g[v]) {
    if (u == from) {
      continue;
    }
    auto tmp = farest(u, v);
    tmp.first++;
    maxi(res, tmp);
  }
  return res;
}
int d1, d2;
vector<int> path;
bool dfs(int v, int from = -1) {
  path.push_back(v);
  if (v == d2) {
    return 1;
  }
  for (int u : g[v]) {
    if (u == from) {
      continue;
    }
    if (dfs(u, v)) {
      return 1;
    }
  }
  path.pop_back();
  return 0;
}
struct Q {
  int a, b, c;
};
long long ans = 0;
vector<Q> res;
int what_diam;
void go(int v, int cur_len, int from = -1) {
  for (int u : g[v]) {
    if (u == from) {
      continue;
    }
    go(u, cur_len + 1, v);
  }
  ans += cur_len;
  res.push_back({v, what_diam, v});
}
void run() {
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    cin >> u >> v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  d1 = farest(1).second;
  d2 = farest(d1).second;
  dfs(d1);
  for (int v : path) {
    in_diam[v] = 1;
  }
  for (int i = 0; i < (int)(path).size(); i++) {
    int v = path[i];
    int len = max(i, (int)(path).size() - i - 1);
    for (int u : g[v]) {
      if (in_diam[u]) {
        continue;
      }
      if (i > (int)(path).size() - i - 1) {
        what_diam = d1;
      } else {
        what_diam = d2;
      }
      go(u, len + 1, v);
    }
  }
  for (int i = 0; i < (int)(path).size() - 1; i++) {
    ans += (int)(path).size() - 1 - i;
    res.push_back({path[i], d2, path[i]});
  }
  cout << ans << "\n";
  for (auto &it : res) {
    cout << it.a << " " << it.b << " " << it.c << "\n";
  }
}