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 = 200010;
struct Edge {
  int v, next;
  Edge(int v = 0, int next = 0) : v(v), next(next) {}
} e[N << 1];
int n, degree[N] = {0};
int id = 0, head[N] = {0};
long long ans = 0;
int st[2], d[2][N] = {0};
inline void addedge(int u, int v) {
  e[++id].v = v;
  e[id].next = head[u];
  head[u] = id;
}
int bfs(int ST, int dis[]) {
  bool vis[N] = {0};
  queue<int> que;
  dis[ST] = 0;
  vis[ST] = true;
  que.push(ST);
  int res = ST;
  while (!que.empty()) {
    int u = que.front();
    que.pop();
    for (int i = head[u]; i; i = e[i].next) {
      int v = e[i].v;
      if (vis[v]) continue;
      res = v;
      vis[v] = true;
      dis[v] = dis[u] + 1;
      que.push(v);
    }
  }
  return res;
}
vector<pair<pair<int, int>, int> > ve;
void update(int u, queue<int>& que) {
  for (int i = head[u]; i; i = e[i].next) {
    int v = e[i].v;
    --degree[v];
    if (degree[v] == 1) que.push(v);
  }
}
void cut() {
  int len = d[0][st[1]];
  queue<int> que;
  for (int i = 1; i <= n; ++i) {
    if (d[0][i] + d[1][i] == len)
      continue;
    else if (degree[i] == 1)
      que.push(i);
  }
  while (!que.empty()) {
    int v = que.front();
    que.pop();
    update(v, que);
    if (d[0][v] > d[1][v]) {
      ans += d[0][v];
      ve.push_back(make_pair(make_pair(v, st[0]), v));
    } else {
      ans += d[1][v];
      ve.push_back(make_pair(make_pair(v, st[1]), v));
    }
  }
}
void dem() {
  int len = d[0][st[1]];
  queue<int> que;
  que.push(st[0]);
  vector<int> de;
  while (!que.empty()) {
    int v = que.front();
    que.pop();
    de.push_back(v);
    update(v, que);
    ans += len;
    --len;
  }
  for (int i = 0; i < (int)de.size(); ++i) {
    ve.push_back(make_pair(make_pair(de[i], st[1]), de[i]));
  }
}
int main() {
  scanf("%d", &n);
  int a, b;
  for (int i = 0; i < n - 1; ++i) {
    scanf("%d%d", &a, &b);
    ++degree[a];
    ++degree[b];
    addedge(a, b);
    addedge(b, a);
  }
  st[0] = bfs(1, d[0]);
  st[1] = bfs(st[0], d[0]);
  bfs(st[1], d[1]);
  cut();
  dem();
  cout << ans << endl;
  for (int i = 0; i < (int)ve.size(); ++i) {
    printf("%d %d %d\n", ve[i].first.first, ve[i].first.second, ve[i].second);
  }
  return 0;
}