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;
inline int rd() {
  int x = 0;
  char ch, t = 0;
  while (!isdigit(ch = getchar())) t |= ch == '-';
  while (isdigit(ch)) x = x * 10 + (ch ^ 48), ch = getchar();
  return x = t ? -x : x;
}
int n, tot, root1, root2, len;
int ds[200005], pre[200005], val[200005], fr[200005];
int hea[200005], nex[200005 * 2], ver[200005 * 2];
long long ans;
bool isdia[200005];
inline void add(int x, int y) {
  ver[++tot] = y, nex[tot] = hea[x], hea[x] = tot;
}
void bfs(int st, int &ed) {
  queue<int> q;
  memset(ds, 0, sizeof(ds)), ds[st] = 0, q.push(st), pre[st] = -1;
  while (!q.empty()) {
    int cur = q.front();
    q.pop();
    for (int i = hea[cur]; i; i = nex[i])
      if (ds[ver[i]] < ds[cur] + 1 && pre[cur] != ver[i])
        ds[ver[i]] = ds[cur] + 1, pre[ver[i]] = cur, q.push(ver[i]);
  }
  for (int i = 1; i <= n; i++)
    if (ds[i] > ds[ed]) ed = i;
}
void dfs1(int x, int dis, int gr) {
  val[x] = dis, fr[x] = gr;
  for (int i = hea[x]; i; i = nex[i]) {
    if (isdia[ver[i]] || val[ver[i]]) continue;
    dfs1(ver[i], dis + 1, gr);
  }
}
void dfs2(int x, int fa, int ro) {
  for (int i = hea[x]; i; i = nex[i]) {
    if (isdia[ver[i]] || ver[i] == fa) continue;
    dfs2(ver[i], x, ro);
  }
  if (fa) printf("%d %d %d\n", ro, x, x);
}
int main() {
  n = rd();
  for (int i = 1, u, v; i < n; i++) u = rd(), v = rd(), add(u, v), add(v, u);
  bfs(1, root1), bfs(root1, root2), len = ds[root2];
  for (int i = root2; i != -1; i = pre[i]) isdia[i] = true;
  for (int i = root2; i != -1; i = pre[i]) dfs1(i, 0, i);
  for (int i = 1; i <= n; i++)
    if (!isdia[i]) ans += val[i] + max(ds[fr[i]], 1 * len - ds[fr[i]]);
  ans += 1ll * len * (len + 1ll) / 2ll;
  printf("%lld\n", ans);
  for (int i = 1; i <= n; i++)
    if (isdia[i]) {
      if (ds[fr[i]] > len - ds[fr[i]])
        dfs2(i, 0, root1);
      else
        dfs2(i, 0, root2);
    }
  for (int i = root2; i != root1; i = pre[i]) printf("%d %d %d\n", root1, i, i);
  return 0;
}