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;
vector<int> g[200222];
char used[200222];
int n, start, finish;
vector<int> d;
void cln() {
  for (int i = 0; i < n; i++) used[i] = 0;
}
pair<int, int> dfs(int v, int len) {
  used[v] = 1;
  pair<int, int> ans;
  ans.first = v, ans.second = len;
  for (int i = 0; i < g[v].size(); i++) {
    int to = g[v][i];
    if (!used[to]) {
      pair<int, int> res = dfs(to, len + 1);
      if (res.second > ans.second) ans = res;
    }
  }
  return ans;
}
bool dfs2(int v) {
  used[v] = 1;
  if (v == finish) return 1;
  for (int i = 0; i < g[v].size(); i++) {
    int to = g[v][i];
    if (!used[to] && dfs2(to)) {
      d.push_back(v);
      return 1;
    }
  }
  return 0;
}
int V, V1, con;
long long L;
vector<pair<int, int> > Ans;
void dfs1(int v, int len) {
  used[v] = 1;
  for (int i = 0; i < g[v].size(); i++) {
    int to = g[v][i];
    if (!used[to]) dfs1(to, len + 1);
  }
  if (V1 == v) return;
  Ans.push_back(pair<int, int>(V, v));
  L += len + con;
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    u--, v--;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  start = dfs(0, 0).first;
  cln();
  finish = dfs(start, 0).first;
  cln();
  d.push_back(finish);
  dfs2(start);
  cln();
  for (int i = 0; i < d.size(); i++) used[d[i]] = 1;
  for (int i = 0; i < d.size(); i++) {
    V1 = d[i];
    con = max(i, (int)d.size() - i - 1);
    if (i > (int)d.size() - i - 1)
      V = d[0];
    else
      V = d.back();
    dfs1(d[i], 0);
  }
  for (int i = 0; i < d.size() - 1; i++) {
    Ans.push_back(pair<int, int>(d.back(), d[i]));
    L += d.size() - i - 1;
  }
  printf("%I64d\n", L);
  for (int i = 0; i < Ans.size(); i++)
    printf("%d %d %d\n", Ans[i].first + 1, Ans[i].second + 1,
           Ans[i].second + 1);
}