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 + 10;
vector<int> g[N];
int dp[N][20], lev[N], cnt[N];
int main() {
  int n;
  cin >> n;
  for (int i = 1; i < n; ++i) {
    int a, b;
    cin >> a >> b;
    g[a].emplace_back(b);
    g[b].emplace_back(a);
  }
  function<int(int, int)> dfs = [&](int s, int p) -> int {
    lev[s] = ~p ? lev[p] + 1 : 0;
    dp[s][0] = ~p ? p : 0;
    int mx = s;
    for (int i : g[s]) {
      if (i == p) continue;
      int res = dfs(i, s);
      if (lev[mx] < lev[res]) mx = res;
      cnt[s]++;
    }
    return mx;
  };
  function<int(int, int)> LCA = [&](int a, int b) -> int {
    if (lev[a] > lev[b]) swap(a, b);
    int diff = lev[b] - lev[a];
    for (int i = 0; i < 20; ++i)
      if (diff >> i & 1) b = dp[b][i];
    if (a == b) return a;
    for (int i = 19; ~i; --i) {
      if (dp[a][i] != dp[b][i]) {
        a = dp[a][i];
        b = dp[b][i];
      }
    }
    return dp[a][0];
  };
  function<int(int, int)> dist = [&](int a, int b) -> int {
    return lev[a] + lev[b] - 2 * lev[LCA(a, b)];
  };
  int p1 = dfs(1, -1);
  memset(cnt, 0, sizeof(cnt));
  int p2 = dfs(p1, -1);
  int cur = p2;
  vector<int> arr;
  set<int> dia;
  while (cur) {
    dia.emplace(cur);
    cur = dp[cur][0];
  }
  for (int j = 1; j < 20; ++j)
    for (int i = 1; i <= n; ++i) dp[i][j] = dp[dp[i][j - 1]][j - 1];
  for (int i = 1; i <= n; ++i) {
    if (dia.count(i)) continue;
    if (!cnt[i]) arr.emplace_back(i);
  }
  long long sum = 0;
  vector<pair<int, int>> ans;
  while (arr.size()) {
    int u = arr.back();
    arr.pop_back();
    int v = dist(u, p1) > dist(u, p2) ? p1 : p2;
    sum += dist(u, v);
    ans.emplace_back(u, v);
    int parent = dp[u][0];
    cnt[parent]--;
    if (!cnt[parent]) arr.emplace_back(parent);
  }
  while (p1 != p2) {
    sum += dist(p1, p2);
    ans.emplace_back(p2, p1);
    p2 = dp[p2][0];
  }
  cout << sum << "\n";
  for (auto i : ans) {
    cout << i.first << " " << i.second << " " << i.first << "\n";
  }
  return 0;
}