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 + 77, Mod = 1e9 + 7;
int n, d1[N], d2[N], P1, P2, Mx, deg[N], p1[N], p2[N];
long long A;
vector<pair<int, pair<int, int> > > R;
vector<int> a[N];
void dfs(int v, int prev, bool t) {
  for (int u : a[v]) {
    if (u == prev) {
      continue;
    }
    if (t) {
      d1[u] = d1[v] + 1;
      p1[u] = v;
    } else {
      d2[u] = d2[v] + 1;
      p2[u] = v;
    }
    dfs(u, v, t);
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i < n; i++) {
    int v, u;
    scanf("%d %d", &v, &u);
    deg[v]++;
    deg[u]++;
    a[v].push_back(u);
    a[u].push_back(v);
  }
  dfs(1, 1, 1);
  for (int i = 1; i <= n; i++) {
    if (d1[i] > Mx) {
      Mx = d1[i];
      P1 = i;
    }
  }
  d1[P1] = 0;
  dfs(P1, P1, 1);
  Mx = 0;
  for (int i = 1; i <= n; i++) {
    if (d1[i] > Mx) {
      Mx = d1[i];
      P2 = i;
    }
  }
  dfs(P2, P2, 0);
  vector<int> Leaves;
  for (int i = 1; i <= n; i++) {
    if (i == P1 || i == P2 || deg[i] > 1) continue;
    Leaves.push_back(i);
  }
  while (Leaves.size()) {
    int v = Leaves.back();
    Leaves.pop_back();
    if (d1[v] >= d2[v]) {
      A += d1[v];
      R.push_back(make_pair(v, make_pair(P1, v)));
    } else {
      A += d2[v];
      R.push_back(make_pair(v, make_pair(P2, v)));
    }
    deg[v]--;
    for (int u : a[v]) {
      deg[u]--;
      if (deg[u] == 1) Leaves.push_back(u);
    }
  }
  while (p2[P1]) {
    R.push_back(make_pair(P1, make_pair(P2, P1)));
    A += d2[P1];
    P1 = p2[P1];
  }
  printf("%I64d\n", A);
  for (auto x : R)
    printf("%d %d %d\n", x.first, x.second.first, x.second.second);
  return 0;
}