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;
inline int read() {
  int X = 0, w = 1;
  char ch = 0;
  while (ch < '0' || ch > '9') {
    if (ch == '-') w = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9')
    X = (X << 3) + (X << 1) + ch - '0', ch = getchar();
  return X * w;
}
const int N = 200018;
vector<int> e[N];
inline void add_edge(const int &u, const int &v) {
  e[u].push_back(v);
  e[v].push_back(u);
}
bool mark[N];
int u, v, disu[N] = {-1}, disv[N] = {-1}, from[N], seq[N];
void dfs1(const int &x, const int &par) {
  disv[x] = disv[par] + 1;
  if (disv[x] > disv[u]) u = x;
  for (unsigned i = 0; i < e[x].size(); i++) {
    const int &y = e[x][i];
    if (y == par) continue;
    dfs1(y, x);
  }
}
void dfs2(const int &x, const int &par) {
  from[x] = par;
  seq[++seq[0]] = x;
  disu[x] = disu[par] + 1;
  if (disu[x] > disu[v]) v = x;
  for (unsigned i = 0; i < e[x].size(); i++) {
    const int &y = e[x][i];
    if (y == par) continue;
    dfs2(y, x);
  }
}
void dfs3(const int &x, const int &par) {
  disv[x] = disv[par] + 1;
  for (unsigned i = 0; i < e[x].size(); i++) {
    const int &y = e[x][i];
    if (y == par) continue;
    dfs3(y, x);
  }
}
struct Answer {
  int a, b, c;
};
Answer ans[N];
int cnt;
int main() {
  const int n = read();
  for (int i = 1; i < n; i++) {
    add_edge(read(), read());
  }
  dfs1(1, 0);
  dfs2(u, 0);
  dfs3(v, 0);
  for (int i = v; i; i = from[i]) mark[i] = 1;
  long long sum = 0;
  for (int i = seq[0]; i; i--) {
    const int x = seq[i];
    if (mark[x]) continue;
    if (disu[x] > disv[x]) {
      ans[cnt++] = (Answer){u, x, x};
      sum += disu[x];
    } else {
      ans[cnt++] = (Answer){v, x, x};
      sum += disv[x];
    }
  }
  for (int i = v; i != u; i = from[i]) {
    ans[cnt++] = (Answer){u, i, i};
    sum += disu[i];
  }
  printf("%lld\n", sum);
  for (int i = 0; i < cnt; i++) {
    printf("%d %d %d\n", ans[i].a, ans[i].b, ans[i].c);
  }
  return 0;
}