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 node {
  int to, next;
} e[500005];
int head[500005], cnt, n, dep[500005], l, r, deep[500005], vis[500005],
    fa[500005], x[500005], y[500005], tot;
long long ans;
void add(int u, int v) {
  cnt++;
  e[cnt].to = v;
  e[cnt].next = head[u];
  head[u] = cnt;
}
void dfs(int u, int f, int kind) {
  for (int i = head[u]; i; i = e[i].next) {
    int v = e[i].to;
    if (v != f) {
      fa[v] = u;
      if (kind == 0)
        dep[v] = dep[u] + 1;
      else
        deep[v] = deep[u] + 1;
      dfs(v, u, kind);
    }
  }
}
void dfss(int u, int f) {
  for (int i = head[u]; i; i = e[i].next) {
    int v = e[i].to;
    if (v != f) {
      dfss(v, u);
    }
  }
  if (!vis[u]) {
    ans += max(dep[u] - dep[l], deep[u] - deep[r]);
    tot++;
    if (dep[u] - dep[l] < deep[u] - deep[r])
      x[tot] = u, y[tot] = r;
    else
      x[tot] = u, y[tot] = l;
  }
}
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);
  }
  dep[1] = 1;
  dfs(1, 0, 0);
  for (int i = 1; i <= n; i++) {
    if (dep[i] > dep[l]) l = i;
  }
  memset(dep, 0, sizeof(dep));
  dep[l] = 1;
  dfs(l, 0, 0);
  for (int i = 1; i <= n; i++) {
    if (dep[i] > dep[r]) r = i;
  }
  deep[r] = 1;
  dfs(r, 0, 1);
  int u = l;
  while (u != r) {
    vis[u] = 1;
    u = fa[u];
  }
  vis[r] = 1;
  dfss(r, 0);
  u = l;
  while (u != r) {
    ans += deep[u] - deep[r];
    tot++;
    x[tot] = u, y[tot] = r;
    u = fa[u];
  }
  cout << ans << endl;
  for (int i = 1; i <= tot; i++) {
    printf("%d %d %d\n", x[i], y[i], x[i]);
  }
  return 0;
}