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;
int n;
int head[200005], nxt[400005], to[400005], tot;
int d[2][200005], p[200005], son[200005];
int a[200005], b[200005], c[200005], sz;
long long ans;
vector<int> v;
void add_edge(int x, int y) {
  tot++;
  to[tot] = y, nxt[tot] = head[x], head[x] = tot;
  return;
}
int bfs(int k, int t) {
  queue<int> q;
  memset(d[t], -1, sizeof d[t]);
  q.push(k);
  d[t][k] = 0;
  p[k] = 0;
  int lst;
  while (!q.empty()) {
    int x = q.front();
    q.pop();
    son[x] = 0;
    for (int i = head[x]; i; i = nxt[i]) {
      int y = to[i];
      if (d[t][y] >= 0) continue;
      q.push(y);
      d[t][y] = d[t][x] + 1;
      p[y] = x;
      son[x]++;
    }
    lst = x;
  }
  return lst;
}
int main() {
  int x, y;
  scanf("%d", &n);
  for (int i = 1; i < n; i++) {
    scanf("%d %d", &x, &y);
    add_edge(x, y);
    add_edge(y, x);
  }
  int nd1 = bfs(1, 0);
  int nd2 = bfs(nd1, 0);
  bfs(nd2, 1);
  for (int i = 1; i <= n; i++)
    if (i != nd1 && !son[i]) v.push_back(i);
  for (int i = 0; i < v.size(); i++) {
    int k = v[i];
    sz++;
    a[sz] = d[1][k] > d[0][k] ? nd2 : nd1;
    b[sz] = c[sz] = k;
    ans += max(d[1][k], d[0][k]);
    son[p[k]]--;
    if (!son[p[k]]) v.push_back(p[k]);
  }
  x = nd1;
  y = d[1][nd1];
  while (y) {
    sz++;
    a[sz] = x, b[sz] = nd2, c[sz] = x;
    ans += y--;
    x = p[x];
  }
  printf("%I64d\n", ans);
  for (int i = 1; i <= sz; i++) printf("%d %d %d\n", a[i], b[i], c[i]);
}