#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 5;
vector<int> e[maxn];
int d[2][maxn];
int c, u, v;
long long ans;
bool vis[maxn];
vector<int> path;
vector<pair<int, int>> ret;
void dfs1(int cur, int fa) {
  for (auto& i : e[cur]) {
    if (i == fa) continue;
    d[0][i] = d[0][cur] + 1;
    if (d[0][i] > d[0][c]) c = i;
    dfs1(i, cur);
  }
}
void dfs2(int cur, int fa) {
  for (auto& i : e[cur]) {
    if (i == fa) continue;
    d[1][i] = d[1][cur] + 1;
    dfs2(i, cur);
    if (i == v) vis[i] = true, path.push_back(cur), v = cur;
  }
}
void dfs3(int cur, int fa) {
  for (auto& i : e[cur]) {
    if (i == fa || vis[i]) continue;
    dfs3(i, cur);
    if (d[0][i] > d[1][i])
      ret.push_back(pair<int, int>(u, i)), ans += d[0][i];
    else
      ret.push_back(pair<int, int>(v, i)), ans += d[1][i];
  }
}
int main() {
  int n, a, b;
  scanf("%d", &n);
  for (int i = 1; i < n; ++i) {
    scanf("%d%d", &a, &b);
    e[a].push_back(b);
    e[b].push_back(a);
  }
  dfs1(1, 0);
  v = u = c;
  d[0][c] = 0, dfs1(c, 0);
  path.push_back(u), vis[c] = true;
  ;
  dfs2(c, 0);
  for (auto& i : path) dfs3(i, 0);
  for (int i = path.size() - 1; i; --i) {
    ret.push_back(pair<int, int>(path[0], path[i]));
    ans += i;
  }
  printf("%lld\n", ans);
  for (auto& i : ret) printf("%d %d %d\n", i.first, i.second, i.second);
}

### Prompt

Create a solution in CPP for the following 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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 5;
vector<int> e[maxn];
int d[2][maxn];
int c, u, v;
long long ans;
bool vis[maxn];
vector<int> path;
vector<pair<int, int>> ret;
void dfs1(int cur, int fa) {
  for (auto& i : e[cur]) {
    if (i == fa) continue;
    d[0][i] = d[0][cur] + 1;
    if (d[0][i] > d[0][c]) c = i;
    dfs1(i, cur);
  }
}
void dfs2(int cur, int fa) {
  for (auto& i : e[cur]) {
    if (i == fa) continue;
    d[1][i] = d[1][cur] + 1;
    dfs2(i, cur);
    if (i == v) vis[i] = true, path.push_back(cur), v = cur;
  }
}
void dfs3(int cur, int fa) {
  for (auto& i : e[cur]) {
    if (i == fa || vis[i]) continue;
    dfs3(i, cur);
    if (d[0][i] > d[1][i])
      ret.push_back(pair<int, int>(u, i)), ans += d[0][i];
    else
      ret.push_back(pair<int, int>(v, i)), ans += d[1][i];
  }
}
int main() {
  int n, a, b;
  scanf("%d", &n);
  for (int i = 1; i < n; ++i) {
    scanf("%d%d", &a, &b);
    e[a].push_back(b);
    e[b].push_back(a);
  }
  dfs1(1, 0);
  v = u = c;
  d[0][c] = 0, dfs1(c, 0);
  path.push_back(u), vis[c] = true;
  ;
  dfs2(c, 0);
  for (auto& i : path) dfs3(i, 0);
  for (int i = path.size() - 1; i; --i) {
    ret.push_back(pair<int, int>(path[0], path[i]));
    ans += i;
  }
  printf("%lld\n", ans);
  for (auto& i : ret) printf("%d %d %d\n", i.first, i.second, i.second);
}

```