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>
int head[200010], to[200010 << 1], next[200010 << 1], d[200010], tot = 0;
inline void $(int u, int v) {
  next[tot] = head[u], to[tot] = v, head[u] = tot++;
  next[tot] = head[v], to[tot] = u, head[v] = tot++;
  d[u]++, d[v]++;
}
int n, dis[200010];
std::queue<int> Q;
inline int fF(int s) {
  memset(dis, -1, sizeof(dis));
  dis[s] = 0;
  register int ret = s;
  Q.push(s);
  while (Q.size()) {
    int x = Q.front();
    Q.pop();
    if (dis[x] > dis[ret]) ret = x;
    for (int i = head[x]; ~i; i = next[i]) {
      if (!~dis[to[i]]) {
        dis[to[i]] = dis[x] + 1;
        Q.push(to[i]);
      }
    }
  }
  return ret;
}
int u, v, spc[200010], fa[200010][18], deep[200010];
int fill(int x, int fa) {
  if (x == v) return spc[x] = 1;
  for (int i = head[x]; ~i; i = next[i]) {
    if (to[i] == fa) continue;
    if (fill(to[i], x)) return spc[x] = 1;
  }
  return 0;
}
void gf(int x) {
  for (int i = head[x]; ~i; i = next[i]) {
    if (to[i] == fa[x][0]) continue;
    fa[to[i]][0] = x;
    deep[to[i]] = deep[x] + 1;
    gf(to[i]);
  }
}
inline int lca(int u, int v) {
  if (deep[u] < deep[v]) std::swap(u, v);
  int delta = deep[u] - deep[v];
  for (int i = 0; i < 18; i++)
    if ((delta >> i) & 1) u = fa[u][i];
  for (int i = 17; i >= 0; i--)
    if (fa[u][i] != fa[v][i]) u = fa[u][i], v = fa[v][i];
  return u == v ? u : fa[u][0];
}
inline int ddd(int u, int v) {
  int l = lca(u, v);
  return deep[u] + deep[v] - (deep[l] << 1);
}
long long ans = 0;
std::vector<int> a, b, c;
inline void shabi() {
  while (Q.size()) {
    int x = Q.front();
    Q.pop();
    ans += std::max(ddd(x, u), ddd(x, v));
    a.push_back(x);
    b.push_back(ddd(x, u) < ddd(x, v) ? v : u);
    c.push_back(x);
    for (int i = head[x]; ~i; i = next[i])
      if (!spc[to[i]] && --d[to[i]] == 1) Q.push(to[i]);
  }
}
void dfs(int x) {
  if (x == v) return;
  a.push_back(x);
  b.push_back(v);
  c.push_back(x);
  ans += ddd(x, v);
  for (int i = head[x]; ~i; i = next[i])
    if (to[i] != fa[x][0] && spc[to[i]]) dfs(to[i]);
}
int main() {
  memset(head, -1, sizeof(head));
  std::cin >> n;
  for (int i = 1; i < n; i++) {
    int u, v;
    std::cin >> u >> v;
    $(u, v);
  }
  v = fF(u = fF(1));
  fill(u, 0);
  gf(u);
  for (int d = 1; d < 18; d++)
    for (int i = 1; i <= n; i++) fa[i][d] = fa[fa[i][d - 1]][d - 1];
  for (int i = 1; i <= n; i++)
    if (d[i] == 1 && !spc[i]) {
      Q.push(i);
    }
  shabi();
  dfs(u);
  std::cout << ans << std::endl;
  for (unsigned i = 0; i < a.size(); i++) {
    std::cout << a[i] << ' ' << b[i] << ' ' << c[i] << std::endl;
  }
  return 0;
}