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 MX = (1 << 20);
vector<int> v[MX];
set<int> S[MX];
int dis[MX], par[MX], depth[MX];
int n;
void Pdfs(int x, int p) {
  for (auto nxt : v[x]) {
    if (nxt == p) continue;
    par[nxt] = x;
    depth[nxt] = depth[x] + 1;
    Pdfs(nxt, x);
  }
}
int bfs(int S, int *fin) {
  for (int j = 1; j <= n; j++) dis[j] = -1;
  queue<int> Q;
  Q.push(S);
  dis[S] = 0;
  while (!Q.empty()) {
    int x = Q.front();
    Q.pop();
    for (auto nxt : v[x]) {
      if (dis[nxt] == -1) {
        dis[nxt] = dis[x] + 1;
        Q.push(nxt);
      }
    }
  }
  int mx = *max_element(dis + 1, dis + 1 + n);
  for (int j = 1; j <= n; j++) fin[j] = dis[j];
  for (int j = 1; j <= n; j++)
    if (dis[j] == mx) return j;
  return 123456789;
}
int is[MX], dis1[MX], dis2[MX];
void getpath(int x, int y) {
  is[x] = is[y] = 1;
  while (x != y) {
    if (depth[x] < depth[y]) swap(x, y);
    x = par[x];
    is[x] = 1;
  }
}
int main() {
  scanf("%d", &n);
  for (int j = 1; j < n; j++) {
    int a, b;
    scanf("%d %d", &a, &b);
    v[a].push_back(b);
    S[a].insert(b);
    v[b].push_back(a);
    S[b].insert(a);
  }
  Pdfs(1, -1);
  int p1 = bfs(1, dis1);
  int p2 = bfs(p1, dis1);
  bfs(p2, dis2);
  getpath(p1, p2);
  long long ans = 0;
  queue<int> Q;
  for (int j = 1; j <= n; j++) {
    if (is[j] == 0 && S[j].size() == 1) Q.push(j);
  }
  vector<pair<int, int> > sol;
  while (!Q.empty()) {
    int x = Q.front();
    Q.pop();
    ans += max(dis1[x], dis2[x]);
    if (dis1[x] >= dis2[x]) {
      sol.push_back({p1, x});
    } else
      sol.push_back({p2, x});
    for (auto nxt : S[x]) {
      S[nxt].erase(x);
      if (S[nxt].size() == 1 && !is[nxt]) {
        Q.push(nxt);
      }
    }
  }
  int diam = dis1[p2];
  ans += 1ll * diam * (diam + 1) / 2;
  while (p1 != p2) {
    if (depth[p1] < depth[p2]) swap(p1, p2);
    sol.push_back({p2, p1});
    p1 = par[p1];
  }
  cout << ans << endl;
  for (auto pp : sol)
    cout << pp.first << ' ' << pp.second << ' ' << pp.second << endl;
}