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;
struct vert {
  int dist, idx, bestU, fromPath;
  bool operator<(const vert& other) const { return dist > other.dist; }
} verts[200009];
int N, best, maxDist, U, V, curU;
vector<int> adjList[200009];
bool dfs(int u, int p, int d) {
  if (d >= maxDist) {
    maxDist = d;
    best = u;
  }
  bool inPath = false;
  if (verts[u].dist < d) {
    verts[u].dist = d;
    verts[u].bestU = curU;
  }
  for (int v : adjList[u]) {
    if (v != p) inPath |= dfs(v, u, d + 1);
  }
  if (u == U) inPath = true;
  verts[u].fromPath = inPath;
  if (inPath) {
    verts[u].dist = d, verts[u].bestU = curU;
  }
  return inPath;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> N;
  int a, b;
  for (int i = 0; i < N - 1; ++i) {
    cin >> a >> b;
    --a, --b;
    adjList[a].push_back(b);
    adjList[b].push_back(a);
  }
  maxDist = 0;
  dfs(0, -1, 0);
  U = curU = best;
  for (int i = 0; i < N; ++i) verts[i] = vert{0, i, -1, 0};
  dfs(U, -1, 0);
  V = curU = best;
  dfs(V, -1, 0);
  sort(verts, verts + N);
  long long int ans = (long long int)maxDist * (maxDist + 1) / 2;
  for (int i = 0; i < N; ++i)
    if (!verts[i].fromPath) ans += verts[i].dist;
  cout << ans << '\n';
  for (int i = 0; i < N; ++i)
    if (!verts[i].fromPath)
      cout << verts[i].idx + 1 << ' ' << verts[i].bestU + 1 << ' '
           << verts[i].idx + 1 << '\n';
  for (int i = 0; i < N; ++i)
    if (verts[i].fromPath && verts[i].dist > 0)
      cout << verts[i].idx + 1 << ' ' << V + 1 << ' ' << verts[i].idx + 1
           << '\n';
}