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 = 2e5 + 5;
struct edge {
  int to, nxt;
} e[N << 1];
int n, dis[N], head[N], cnt = 0, u, v, fa[N];
long long ans = 0;
bool on[N];
vector<pair<int, int>> s;
void dfs(int now) {
  for (int i = head[now]; i; i = e[i].nxt)
    if (!~dis[e[i].to]) {
      dis[e[i].to] = dis[now] + 1;
      dfs(e[i].to);
    }
}
void dfs2(int now) {
  for (int i = head[now]; i; i = e[i].nxt)
    if (dis[e[i].to] > dis[now]) {
      fa[e[i].to] = now;
      dfs2(e[i].to);
      on[now] = on[now] || on[e[i].to];
    }
}
inline void dfs3(int now, int rt) {
  for (int i = head[now]; i; i = e[i].nxt)
    if (dis[e[i].to] > dis[now]) dfs3(e[i].to, on[e[i].to] ? e[i].to : rt);
  if (!on[now]) {
    if (dis[now] > dis[now] + dis[v] - (dis[rt] << 1)) {
      ans += dis[now];
      s.push_back(make_pair(u, now));
    } else {
      ans += dis[now] + dis[v] - (dis[rt] << 1);
      s.push_back(make_pair(v, now));
    }
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i < n; ++i) {
    scanf("%d%d", &u, &v);
    e[++cnt] = (edge){v, head[u]};
    head[u] = cnt;
    e[++cnt] = (edge){u, head[v]};
    head[v] = cnt;
  }
  memset(dis, -1, sizeof dis);
  dis[1] = 0;
  dfs(1);
  u = 1;
  for (int i = 2; i <= n; ++i)
    if (dis[i] > dis[u]) u = i;
  memset(dis, -1, sizeof dis);
  dis[u] = 0;
  dfs(u);
  v = u;
  for (int i = 1; i <= n; ++i)
    if (dis[v] < dis[i]) v = i;
  on[v] = true;
  dfs2(u);
  dfs3(u, u);
  for (; u != v; v = fa[v]) ans += dis[v], s.push_back(make_pair(u, v));
  printf("%lld\n", ans);
  for (auto i : s) printf("%d %d %d\n", i.first, i.second, i.second);
  return 0;
}