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 N = 2e5 + 10;
vector<int> v[N];
int n;
int cnt;
int x, y;
int L, R;
int f[N][19];
int a[N];
int deep[N], father[N];
int Father[N];
int vis[N];
int c[N];
int isroot[N];
long long ans = 0;
queue<int> q;
void dfs(int x, int fa, int now) {
  deep[x] = now;
  if (fa) {
    f[x][0] = fa;
    for (int i = 0; f[f[x][i]][i]; ++i) f[x][i + 1] = f[f[x][i]][i];
  }
  if (now > cnt) cnt = now, L = x;
  for (auto u : v[x]) {
    if (u == fa) continue;
    dfs(u, x, now + 1);
  }
}
void dfs2(int x, int fa, int now) {
  father[x] = fa;
  if (now > cnt) cnt = now, R = x;
  for (auto u : v[x]) {
    if (u == fa) continue;
    dfs2(u, x, now + 1);
  }
}
int LCA(int a, int b) {
  if (deep[a] < deep[b]) swap(a, b);
  for (int i = 0, delta = deep[a] - deep[b]; delta; delta >>= 1, ++i)
    if (delta & 1) a = f[a][i];
  if (a == b) return a;
  for (int i(18); i >= (0); --i)
    if (f[a][i] != f[b][i]) {
      a = f[a][i];
      b = f[b][i];
    }
  return f[a][0];
}
int dis(int x, int y) {
  int z = LCA(x, y);
  return deep[x] + deep[y] - 2 * deep[z];
}
void dfs3(int x, int fa) {
  vis[x] = 1;
  Father[x] = fa;
  for (auto u : v[x]) {
    if (vis[u]) continue;
    dfs3(u, x);
    ++c[x];
  }
}
int main() {
  scanf("%d", &n);
  for (int i(1); i <= (n); ++i) {
    scanf("%d%d", &x, &y);
    v[x].push_back(y);
    v[y].push_back(x);
  }
  L = 0, cnt = 0;
  dfs(1, 0, 0);
  cnt = 0;
  R = 0;
  memset(father, 0, sizeof father);
  dfs2(L, 0, 0);
  cnt = 0;
  x = R;
  while (x) {
    ++cnt;
    a[cnt] = x;
    x = father[x];
  }
  memset(vis, 0, sizeof vis);
  for (int i(1); i <= (cnt); ++i) vis[a[i]] = 1;
  for (int i(1); i <= (n); ++i) isroot[i] = vis[i];
  for (int i(1); i <= (n); ++i)
    if (!vis[i]) {
      int now = max(dis(i, L), dis(i, R));
      ans += 1ll * now;
    }
  memset(c, 0, sizeof c);
  for (int i(1); i <= (cnt); ++i) dfs3(a[i], 0);
  for (int i(1); i <= (n); ++i)
    if (!isroot[i] && c[i] == 0) q.push(i);
  ans = ans + 1ll * cnt * (cnt - 1) / 2;
  printf("%lld\n", ans);
  while (!q.empty()) {
    x = q.front();
    q.pop();
    vis[x] = 1;
    int al = dis(x, L), ar = dis(x, R);
    if (al > ar)
      printf("%d %d %d\n", L, x, x);
    else
      printf("%d %d %d\n", R, x, x);
    int u = Father[x];
    --c[u];
    if (c[u] == 0 && !isroot[u]) q.push(u);
  }
  for (int i(1); i <= (cnt - 1); ++i) printf("%d %d %d\n", a[i], a[cnt], a[i]);
  return 0;
}