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>
inline int getint() {
  register int num = 0, sgn = 1;
  register char ch = getchar();
  for (; !isdigit(ch); ch = getchar())
    if (ch == '-') sgn = 0;
  for (; isdigit(ch); ch = getchar()) num = (num << 3) + (num << 1) + ch - '0';
  return sgn ? num : -num;
}
int head[200002], next[200002 << 1], to[200002 << 1], tot = 0;
inline void addedge(int u, int v) {
  next[++tot] = head[u];
  to[tot] = v;
  head[u] = tot;
  next[++tot] = head[v];
  to[tot] = u;
  head[v] = tot;
}
bool mark[200002];
int u, v, n, disu[200002] = {-1}, disv[200002] = {-1}, fa[200002], seq[200002];
void dfs1(int x, int f) {
  disv[x] = disv[f] + 1;
  if (disv[x] > disv[u]) u = x;
  for (register int i = head[x], y = to[i]; ~i; y = to[i = next[i]])
    if (y != f) dfs1(y, x);
}
void dfs2(int x, int f) {
  fa[x] = f;
  seq[++seq[0]] = x;
  disu[x] = disu[f] + 1;
  if (disu[x] > disu[v]) v = x;
  for (register int i = head[x], y = to[i]; ~i; y = to[i = next[i]])
    if (y != f) dfs2(y, x);
}
void dfs3(int x, int f) {
  disv[x] = disv[f] + 1;
  for (register int i = head[x], y = to[i]; ~i; y = to[i = next[i]])
    if (y != f) dfs3(y, x);
}
struct answer {
  int a, b, c;
} ans[200002];
int cnt;
int main() {
  memset(head, -1, sizeof(head));
  n = getint();
  for (register int i = 1; i < n; i++) addedge(getint(), getint());
  dfs1(1, 0);
  dfs2(u, 0);
  dfs3(v, 0);
  for (register int i = v; i; i = fa[i]) mark[i] = 1;
  long long sum = 0;
  for (register int i = seq[0]; i; i--) {
    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 (register int i = v; i != u; i = fa[i]) {
    ans[++cnt] = (answer){u, i, i};
    sum += disu[i];
  }
  printf("%I64d\n", sum);
  for (register int i = 1; i <= cnt; i++)
    printf("%d %d %d\n", ans[i].a, ans[i].b, ans[i].c);
  return 0;
}