#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 5;
vector<int> G[maxn];
vector<pair<int, int>> a;
int dep[maxn], f[maxn], in[maxn];
int dia, d;
long long ans;
void dfs(int u, int fa) {
  dep[u] = dep[f[u] = fa] + 1;
  if (dep[u] > dia) dia = dep[u], d = u;
  for (auto i : G[u])
    if (i - fa) dfs(i, u);
}
void dfs2(int u, int fa, int dis, int flg = 0) {
  in[u] = 1;
  for (auto i : G[u])
    if (!in[i]) dfs2(i, fa, dis + 1, 1);
  if (flg) ans += dis, a.push_back(make_pair(fa, u));
}
int main() {
  int n, x, y;
  scanf("%d", &n);
  for (int i = 1; i < n; ++i)
    scanf("%d%d", &x, &y), G[x].push_back(y), G[y].push_back(x);
  dfs(1, 0);
  x = d;
  dia = 0, dfs(x, 0);
  y = d;
  ans = (long long)(dia - 1) * dia / 2;
  for (int i = y; i; i = f[i]) in[i] = 1;
  for (int i = y; i; i = f[i])
    if (dep[i] - 1 > dep[y] - dep[i])
      dfs2(i, x, dep[i] - 1);
    else
      dfs2(i, y, dep[y] - dep[i]);
  printf("%lld\n", ans);
  for (auto i : a) printf("%d %d %d\n", i.first, i.second, i.second);
  for (int i = y; i - x; i = f[i]) printf("%d %d %d\n", x, i, i);
}

### Prompt

Your task is to create a Cpp solution to 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> G[maxn];
vector<pair<int, int>> a;
int dep[maxn], f[maxn], in[maxn];
int dia, d;
long long ans;
void dfs(int u, int fa) {
  dep[u] = dep[f[u] = fa] + 1;
  if (dep[u] > dia) dia = dep[u], d = u;
  for (auto i : G[u])
    if (i - fa) dfs(i, u);
}
void dfs2(int u, int fa, int dis, int flg = 0) {
  in[u] = 1;
  for (auto i : G[u])
    if (!in[i]) dfs2(i, fa, dis + 1, 1);
  if (flg) ans += dis, a.push_back(make_pair(fa, u));
}
int main() {
  int n, x, y;
  scanf("%d", &n);
  for (int i = 1; i < n; ++i)
    scanf("%d%d", &x, &y), G[x].push_back(y), G[y].push_back(x);
  dfs(1, 0);
  x = d;
  dia = 0, dfs(x, 0);
  y = d;
  ans = (long long)(dia - 1) * dia / 2;
  for (int i = y; i; i = f[i]) in[i] = 1;
  for (int i = y; i; i = f[i])
    if (dep[i] - 1 > dep[y] - dep[i])
      dfs2(i, x, dep[i] - 1);
    else
      dfs2(i, y, dep[y] - dep[i]);
  printf("%lld\n", ans);
  for (auto i : a) printf("%d %d %d\n", i.first, i.second, i.second);
  for (int i = y; i - x; i = f[i]) printf("%d %d %d\n", x, i, i);
}

```