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 = 200010;
int len, u, v;
long long nAns = 0;
vector<int> g[maxn];
int d[maxn], rt[maxn];
bool dm[maxn];
pair<int, int> pAns[maxn];
void dfsD(int id, int fa) {
  int i, nxt;
  for (i = 0; i < g[id].size(); i++) {
    nxt = g[id][i];
    if (nxt == fa) continue;
    rt[nxt] = id;
    d[nxt] = d[id] + 1;
    dfsD(nxt, id);
  }
}
void dfs(int id, int di) {
  int i, nxt;
  for (i = 0; i < g[id].size(); i++) {
    nxt = g[id][i];
    if (d[nxt] <= d[id]) continue;
    if (dm[nxt])
      dfs(nxt, nxt);
    else
      dfs(nxt, di);
  }
  if (!dm[id]) {
    if (d[di] * 2 > d[v]) {
      nAns += d[id];
      pAns[++len] = make_pair(u, id);
    } else {
      nAns += d[id] + d[v] - d[di] * 2;
      pAns[++len] = make_pair(v, id);
    }
  }
}
int main() {
  int i, n;
  scanf("%d", &n);
  for (i = 1; i < n; i++) {
    scanf("%d%d", &u, &v);
    g[u].push_back(v);
    g[v].push_back(u);
  }
  dfsD(1, 0);
  for (i = u = 1; i <= n; i++)
    if (d[i] > d[u]) u = i;
  d[u] = 0;
  dfsD(u, 0);
  for (i = v = 1; i <= n; i++)
    if (d[i] > d[v]) v = i;
  for (i = v; i != u; i = rt[i]) dm[i] = true;
  dm[u] = true;
  dfs(u, u);
  for (i = v; i != u; i = rt[i]) {
    nAns += d[i];
    pAns[++len] = make_pair(u, i);
  }
  printf("%lld\n", nAns);
  for (i = 1; i <= len; i++)
    printf("%d %d %d\n", pAns[i].first, pAns[i].second, pAns[i].second);
  return 0;
}