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;
struct edge {
  int to, next;
} e[200005 * 2];
int d[200005], dl[200005], dr[200005], dd[200005];
int head[200005], q[200005], id[200005], a[200005];
long long ans;
int n, tot;
bool cmp(int x, int y) {
  if (dd[x] != dd[y]) return dd[x] > dd[y];
  return dl[x] > dl[y];
}
void add(int x, int y) {
  e[++tot] = (edge){y, head[x]};
  head[x] = tot;
}
void bfs(int S, int *d) {
  for (int i = 1; i <= n; i++) d[i] = -1;
  d[S] = 0;
  q[1] = S;
  for (int h = 0, t = 1; h != t;)
    for (int x = q[++h], i = head[x]; i; i = e[i].next)
      if (d[e[i].to] == -1) d[e[i].to] = d[x] + a[e[i].to], q[++t] = e[i].to;
}
int findmax(int *d) {
  int mx = 1;
  for (int i = 2; i <= n; i++)
    if (d[mx] < d[i]) mx = i;
  return mx;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i < n; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    add(x, y);
    add(y, x);
  }
  for (int i = 1; i <= n; i++) a[i] = 1;
  bfs(1, d);
  int l = findmax(d);
  bfs(l, dl);
  int r = findmax(dl);
  bfs(r, dr);
  for (int i = 1; i <= n; i++)
    if (dl[i] + dr[i] == dl[r])
      a[i] = 0;
    else
      ans += max(dl[i], dr[i]);
  ans += 1ll * dl[r] * (dl[r] + 1) / 2;
  printf("%I64d\n", ans);
  bfs(l, dd);
  for (int i = 1; i <= n; i++) id[i] = i;
  sort(id + 1, id + n + 1, cmp);
  for (int i = 1; i < n; i++)
    if (dd[id[i]]) {
      int tmp = dl[id[i]] > dr[id[i]] ? l : r;
      printf("%d %d %d\n", tmp, id[i], id[i]);
    } else
      printf("%d %d %d\n", l, id[i], id[i]);
}