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 f = 1, int x = 0, char ch = ' ') {
  while (!isdigit(ch = getchar()))
    if (ch == '-') f = -1;
  while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
  return f * x;
}
const int N = 2e5 + 5;
struct Edge {
  int next, to;
  Edge(int next = 0, int to = 0) : next(next), to(to){};
} edge[N << 1];
int tot, head[N];
void _add(int x, int y) { edge[++tot] = Edge(head[x], y), head[x] = tot; }
void add(int x, int y) { _add(x, y), _add(y, x); }
int n, m, d[N], fa[N], path[N], u, v;
void dfs(int x, int f) {
  fa[x] = f, d[x] = d[f] + 1;
  for (int i = head[x]; i; i = edge[i].next) {
    int y = edge[i].to;
    if (y == f) continue;
    dfs(y, x);
  }
}
int meth[3][N];
long long ans;
void solve(int x, int topf, int w) {
  if (path[x]) topf = x, w = d[x] - d[u] > d[v] - d[x] ? u : v;
  for (int i = head[x]; i; i = edge[i].next) {
    int y = edge[i].to;
    if (y == fa[x]) continue;
    solve(y, topf, w);
  }
  if (!path[x])
    meth[0][++m] = x, meth[1][m] = w, meth[2][m] = x,
    ans += d[x] - d[topf] + abs(d[topf] - d[w]);
}
int main() {
  n = read();
  for (int i = 1, x, y; i < n; ++i) x = read(), y = read(), add(x, y);
  dfs(1, 0);
  for (int i = 1; i <= n; ++i)
    if (d[i] > d[u]) u = i;
  dfs(u, 0);
  for (int i = 1; i <= n; ++i)
    if (d[i] > d[v]) v = i;
  for (int x = v; x; x = fa[x]) path[x] = 1;
  solve(u, 0, 0);
  for (int x = v; x != u; x = fa[x])
    meth[0][++m] = x, meth[1][m] = u, meth[2][m] = x, ans += d[x] - 1;
  printf("%lld\n", ans);
  for (int i = 1; i < n; ++i, puts(""))
    for (int k = 0; k < 3; ++k) printf("%d ", meth[k][i]);
  return 0;
}