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;
vector<int> G[maxn];
int d[2][maxn], p[maxn], deg[maxn];
void dfs(int now, int fa, int dep, int idx) {
  if (~idx) d[idx][now] = dep;
  p[now] = fa;
  for (int u : G[now])
    if (u != fa) dfs(u, now, dep + 1, idx);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  for (int i = 0; i < n - 1; ++i) {
    int a, b;
    cin >> a >> b;
    G[a].push_back(b);
    G[b].push_back(a);
  }
  dfs(1, 0, 0, 0);
  int bst = -1;
  for (int i = 1; i <= n; ++i) {
    if (bst == -1 || d[0][i] > d[0][bst]) bst = i;
  }
  dfs(bst, 0, 0, 0);
  int to = -1;
  for (int i = 1; i <= n; ++i) {
    if (to == -1 || d[0][i] > d[0][to]) to = i;
  }
  dfs(to, 0, 0, 1);
  int diameter = d[1][bst], now = to;
  for (int i = 0; i < diameter / 2; ++i) now = p[now];
  int center = now;
  dfs(center, 0, 0, -1);
  queue<int> leaf;
  for (int i = 1; i <= n; ++i) {
    if (i == bst || i == to) continue;
    if (G[i].size() == 1 && G[i][0] == p[i])
      leaf.push(i);
    else
      deg[i] = G[i].size() - 1;
  }
  long long ans = 0;
  vector<tuple<int, int, int>> vec;
  int ver[2] = {bst, to};
  while (leaf.size()) {
    int now = leaf.front();
    leaf.pop();
    int idx = (d[0][now] > d[1][now] ? 0 : 1);
    ans += d[idx][now];
    vec.emplace_back(ver[idx], now, now);
    for (int u : G[now]) {
      --deg[u];
      if (deg[u] == 0) leaf.push(u);
    }
  }
  dfs(to, 0, 0, 1);
  now = bst;
  for (now; now != to; now = p[now]) {
    ans += diameter;
    --diameter;
    vec.emplace_back(now, to, now);
  }
  cout << ans << endl;
  assert(vec.size() == n - 1);
  for (auto i : vec) {
    int a, b, c;
    tie(a, b, c) = i;
    cout << a << ' ' << b << ' ' << c << endl;
  }
  return 0;
}