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 maxn = 2e5 + 5;
int n, len;
int d[maxn][2];
vector<int> from[maxn];
int q[maxn];
int lay[maxn], deg[maxn];
long long ans;
vector<pair<int, int> > p;
void dfs(int u, int last, int t) {
  d[u][t] = (last == 0 ? -1 : d[last][t]) + 1;
  for (auto v : from[u])
    if (v != last) dfs(v, u, t);
}
void dfs2(int u, int last, int st) {
  for (auto v : from[u])
    if (v != last && lay[v]) dfs2(v, u, st);
  ans += d[u][0];
  if (st != u) p.push_back({st, u});
}
int main() {
  int i, x, y, l, r, u, v;
  scanf("%d", &n);
  for (i = 0; i < n - 1; i++) {
    scanf("%d%d", &x, &y);
    from[x].push_back(y);
    from[y].push_back(x);
  }
  dfs(1, 0, 0);
  pair<int, int> t1 = {-1, 0};
  for (i = 1; i <= n; i++) t1 = max(t1, {d[i][0], i});
  dfs(t1.second, 0, 0);
  pair<int, int> t2 = {-1, 0};
  for (i = 1; i <= n; i++) t2 = max(t2, {d[i][0], i});
  dfs(t2.second, 0, 1);
  len = t2.first;
  l = 1;
  r = 0;
  for (i = 1; i <= n; i++)
    deg[i] = from[i].size(), lay[i] = (d[i][0] + d[i][1] == len);
  for (i = 1; i <= n; i++)
    if (deg[i] == 1 && !lay[i]) q[++r] = i;
  while (l <= r) {
    u = q[l++];
    if (d[u][0] >= d[u][1])
      p.push_back({t1.second, u}), ans += d[u][0];
    else
      p.push_back({t2.second, u}), ans += d[u][1];
    for (auto v : from[u]) {
      deg[v]--;
      if (deg[v] == 1 && !lay[v]) q[++r] = v;
    }
  }
  dfs2(t1.second, 0, t1.second);
  printf("%I64d\n", ans);
  for (auto t : p) printf("%d %d %d\n", t.first, t.second, t.second);
}