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 = 200000 + 10;
const int INF = 0x3f3f3f3f;
vector<int> g[maxn], rk[maxn];
int d[maxn], du[maxn], dv[maxn], p[maxn], used[maxn];
void bfs(int s, int n, int* dis) {
  queue<int> q;
  for (int i = 1; i <= n; i++) dis[i] = INF;
  dis[s] = 0;
  q.push(s);
  while (!q.empty()) {
    int u = q.front();
    q.pop();
    for (int i = 0; i < g[u].size(); i++) {
      int to = g[u][i];
      if (dis[to] != INF) continue;
      dis[to] = dis[u] + 1;
      p[to] = u;
      q.push(to);
    }
  }
}
int main() {
  int n, a, b;
  scanf("%d", &n);
  for (int i = 0; i < n - 1; i++) {
    scanf("%d%d", &a, &b);
    g[a].push_back(b);
    g[b].push_back(a);
  }
  bfs(1, n, d);
  int u, v, temp = 0;
  for (int i = 1; i <= n; i++)
    if (d[i] > temp) {
      u = i;
      temp = d[i];
    }
  bfs(u, n, du);
  temp = 0;
  for (int i = 1; i <= n; i++)
    if (du[i] > temp) {
      v = i;
      temp = du[i];
    }
  bfs(v, n, dv);
  temp = u;
  memset(used, 0, sizeof used);
  while (temp != v) {
    used[temp] = 1;
    temp = p[temp];
  }
  used[v] = 1;
  long long ans = 0;
  for (int i = 1; i <= n; i++)
    if (!used[i]) {
      rk[dv[i]].push_back(i);
    }
  for (int i = dv[u]; i >= 1; i--) {
    for (int j = 0; j < rk[i].size(); j++)
      if (!used[rk[i][j]]) {
        int tt = rk[i][j];
        ans += max(du[tt], dv[tt]);
      }
  }
  ans += 1ll * dv[u] * (dv[u] + 1) / 2;
  printf("%I64d\n", ans);
  for (int i = dv[u]; i >= 1; i--) {
    for (int j = 0; j < rk[i].size(); j++)
      if (!used[rk[i][j]]) {
        int tt = rk[i][j];
        if (du[tt] > dv[tt])
          printf("%d %d %d\n", u, tt, tt);
        else
          printf("%d %d %d\n", v, tt, tt);
      }
  }
  temp = u;
  while (temp != v) {
    printf("%d %d %d\n", temp, v, temp);
    temp = p[temp];
  }
  return 0;
}