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 MAXN = 200005;
vector<int> e[MAXN];
int dis[MAXN], key[MAXN], fa[MAXN][18];
void dfs(int u, int f) {
  fa[u][0] = f;
  for (int i = 1; i < 18; i++) fa[u][i] = fa[fa[u][i - 1]][i - 1];
  for (auto &v : e[u]) {
    if (v == f) continue;
    dis[v] = dis[u] + 1;
    dfs(v, u);
  }
}
inline int lca(int u, int v) {
  if (dis[u] < dis[v]) swap(u, v);
  for (int i = 0, d = dis[u] - dis[v]; d; i++, d >>= 1)
    if (d & 1) u = fa[u][i];
  if (u == v) return u;
  for (int i = 17; i >= 0; i--)
    if (fa[u][i] != fa[v][i]) u = fa[u][i], v = fa[v][i];
  return fa[u][0];
}
inline int len(int u, int v) { return dis[u] + dis[v] - 2 * dis[lca(u, v)]; }
vector<int> dia;
bool get(int u, int f, int tar) {
  if (u == tar) return 1;
  for (auto &v : e[u]) {
    if (v == f) continue;
    if (get(v, u, tar)) {
      dia.push_back(v);
      return 1;
    }
  }
  return 0;
}
vector<pair<int, int> > opt;
void build(int u, int f, int lef, int rig, long long &res) {
  for (auto &v : e[u]) {
    if (v == f || key[v]) continue;
    int chs;
    if (len(lef, v) > len(rig, v))
      chs = lef;
    else
      chs = rig;
    res += len(chs, v);
    opt.push_back(make_pair(chs, v));
    build(v, u, lef, rig, res);
  }
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n - 1; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    e[u].push_back(v);
    e[v].push_back(u);
  }
  dfs(1, dis[1] = 0);
  int u = 1;
  for (int i = 1; i <= n; i++)
    if (dis[i] > dis[u]) u = i;
  dfs(u, dis[u] = 0);
  int v = 1;
  for (int i = 1; i <= n; i++)
    if (dis[i] > dis[v]) v = i;
  get(u, 0, v);
  dia.push_back(u);
  long long res = 0;
  for (int i = 0; i < (int)dia.size(); i++) {
    key[dia[i]] = 1, res += i;
    if (i) opt.push_back(make_pair(dia[0], dia[i]));
  }
  for (int i = 0; i < (int)dia.size(); i++) build(dia[i], 0, u, v, res);
  printf("%lld\n", res);
  reverse(opt.begin(), opt.end());
  for (int i = 0; i < (int)opt.size(); i++)
    printf("%d %d %d\n", opt[i].first, opt[i].second, opt[i].second);
  return 0;
}