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 N = 200005;
struct _node {
  int to, next;
} num[N * 2];
struct node {
  int x, y, z;
  node() {}
  node(int a, int b, int c) {
    x = a;
    y = b;
    z = c;
  }
} ans[N];
int cnt, head[N], n, u, v, dep[N], f[N], mark[N], tot, rt, tt;
long long sum;
void add(int x, int y) {
  num[++cnt].to = y;
  num[cnt].next = head[x];
  head[x] = cnt;
}
void dfs(int x, int fa) {
  for (int i = head[x]; i; i = num[i].next)
    if (num[i].to != fa) {
      dep[num[i].to] = dep[x] + 1;
      f[num[i].to] = x;
      dfs(num[i].to, x);
    }
}
void solve(int x, int fa, int d) {
  if (mark[x]) d++;
  for (int i = head[x]; i; i = num[i].next)
    if (num[i].to != fa) {
      solve(num[i].to, x, d);
      if (mark[num[i].to]) continue;
      int x1 = dep[num[i].to], x2 = dep[tt] + dep[num[i].to] - 2 * d;
      if (x1 >= x2)
        sum += x1, ans[++tot] = node(rt, num[i].to, num[i].to);
      else
        sum += x2, ans[++tot] = node(tt, num[i].to, num[i].to);
    }
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i < n; i++) scanf("%d%d", &u, &v), add(u, v), add(v, u);
  dfs(1, 0);
  for (int i = 1; i <= n; i++)
    if (dep[i] > dep[rt]) rt = i;
  dep[rt] = 0;
  f[rt] = 0;
  dfs(rt, 0);
  for (int i = 1; i <= n; i++)
    if (dep[i] > dep[tt]) tt = i;
  for (int i = tt; i != rt; i = f[i]) mark[i] = 1;
  solve(rt, 0, 0);
  for (int i = tt; i != rt; i = f[i])
    ans[++tot] = node(rt, i, i), sum += dep[i];
  printf("%lld\n", sum);
  for (int i = 1; i <= tot; i++)
    printf("%d %d %d\n", ans[i].x, ans[i].y, ans[i].z);
  return 0;
}