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 p3i {
  int u, v, c;
};
const int maxn = 2e5 + 10;
vector<int> G[maxn];
bool vis[maxn];
int deg[maxn];
int par[maxn];
int tmp[maxn];
int dist0[maxn];
int distf[maxn];
int distl[maxn];
vector<p3i> mf;
int fd, ld;
long long ans = 0;
inline void bfs(int source, int dist[]) {
  memset(vis, 0, sizeof vis);
  queue<int> bq;
  vis[source] = true;
  dist[source] = 0;
  par[source] = source;
  bq.push(source);
  while (!bq.empty()) {
    int t = bq.front();
    bq.pop();
    for (int u : G[t]) {
      if (vis[u]) continue;
      vis[u] = true;
      dist[u] = dist[t] + 1;
      par[u] = t;
      bq.push(u);
    }
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  for (int i = 1; i < n; i++) {
    int x, y;
    cin >> x >> y;
    x--;
    y--;
    G[x].push_back(y);
    G[y].push_back(x);
    deg[x]++;
    deg[y]++;
  }
  bfs(0, dist0);
  fd = max_element(dist0, dist0 + n) - dist0;
  bfs(fd, distf);
  ld = max_element(distf, distf + n) - distf;
  bfs(ld, distl);
  queue<int> lfs;
  for (int v = 0; v < n; v++)
    if (deg[v] == 1) lfs.push(v);
  memset(vis, 0, sizeof vis);
  while (!lfs.empty()) {
    int t = lfs.front();
    lfs.pop();
    if (t == fd or t == ld) continue;
    vis[t] = true;
    int x = par[t];
    if (--deg[x] == 1) lfs.push(x);
    int cand = (distf[t] > distl[t] ? fd : ld);
    mf.push_back({t, cand, t});
    ans += max(distf[t], distl[t]);
  }
  int cur = fd;
  while (cur != ld) {
    mf.push_back({cur, ld, cur});
    ans += distl[cur];
    cur = par[cur];
  }
  cout << ans << endl;
  for (auto x : mf) cout << x.u + 1 << " " << x.v + 1 << " " << x.c + 1 << endl;
  return 0;
}