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;
int n, m, i, s, j, u, v, head[200005], Next[400005], adj[400005], k,
    siz[200005], mxx, l, nx[200005], a[200005], p;
long long ans, mx;
bool flag = true;
void Push(int u, int v) {
  Next[++k] = head[u];
  head[u] = k;
  adj[k] = v;
}
void dfs(int i, int f, int d) {
  int j;
  if (mx < d) {
    mx = d;
    mxx = i;
  }
  for (j = head[i]; j != 0; j = Next[j])
    if (adj[j] != f) dfs(adj[j], i, d + 1);
}
void dfs2(int i, int f) {
  int j;
  if (i == mxx) {
    nx[i] = -1;
    flag = false;
    return;
  }
  for (j = head[i]; j != 0; j = Next[j])
    if (adj[j] != f && flag) {
      nx[i] = adj[j];
      dfs2(adj[j], i);
    }
  if (flag) nx[i] = 0;
}
void dfs3(int i, int f, int d) {
  int j;
  if (nx[i] == 0) ans += d;
  for (j = head[i]; j != 0; j = Next[j])
    if (adj[j] != f && adj[j] != nx[i])
      if (nx[i] != 0)
        dfs3(adj[j], i, max(1ll * d, mx - d) + 1);
      else
        dfs3(adj[j], i, d + 1);
  if (nx[i] != 0 && nx[i] != -1) dfs3(nx[i], i, d + 1);
}
void dfs4(int i, int f, int tot) {
  int j;
  for (j = head[i]; j != 0; j = Next[j])
    if (adj[j] != f) dfs4(adj[j], i, tot);
  printf("%d %d %d\n", tot, i, i);
}
int main() {
  scanf("%d", &n);
  for (i = 1; i < n; i++) {
    scanf("%d %d", &u, &v);
    Push(u, v);
    Push(v, u);
  }
  mx = -1;
  dfs(1, 0, 0);
  l = mxx;
  mx = -1;
  dfs(mxx, 0, 0);
  dfs2(l, 0);
  dfs3(l, i, 0);
  cout << ans + mx * (mx + 1) / 2 << endl;
  for (i = l; i != -1; i = nx[i]) a[++p] = i;
  for (i = 1; i <= p; i++)
    for (j = head[a[i]]; j != 0; j = Next[j])
      if (adj[j] != a[i - 1] && adj[j] != a[i + 1])
        if (i <= (mx + 1) / 2)
          dfs4(adj[j], a[i], mxx);
        else
          dfs4(adj[j], a[i], l);
  for (i = p; i > 1; i--) printf("%d %d %d\n", l, a[i], a[i]);
}