#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const int N = 2e5 + 5;
struct P {
  int x, y, z;
};
int n, mx, id, zy, yh, d[2][N], isd[N];
ll ans;
basic_string<int> G[N];
basic_string<P> res;
queue<int> q;
void dfs(int u, int fa, int d) {
  if (d > mx) mx = d, id = u;
  for (int v : G[u])
    if (v != fa) dfs(v, u, d + 1);
}
int mark(int u, int fa) {
  isd[u] = 1;
  if (u == yh) return 1;
  for (int v : G[u])
    if (v != fa && mark(v, u)) return 1;
  return isd[u] = 0;
}
void getd(int u, int fa, int f) {
  for (int v : G[u])
    if (v != fa) {
      if (f) {
        if (isd[v])
          res += {u, zy, u}, ans += d[0][u];
        else
          q.push(v);
      }
      d[f][v] = d[f][u] + 1, getd(v, u, f);
    }
}
int main() {
  scanf("%d", &n);
  for (int i = 1, a, b; i < n; ++i) scanf("%d%d", &a, &b), G[a] += b, G[b] += a;
  dfs(1, 0, 0), zy = id, mx = 0, dfs(id, 0, 0), yh = id;
  mark(zy, 0), getd(zy, 0, 0), getd(yh, 0, 1);
  reverse(res.begin(), res.end());
  while (!q.empty()) {
    int u = q.front();
    q.pop();
    if (d[0][u] > d[1][u])
      res += {u, zy, u}, ans += d[0][u];
    else
      res += {u, yh, u}, ans += d[1][u];
  }
  printf("%lld\n", ans);
  for (auto it = res.rbegin(); it != res.rend(); ++it)
    printf("%d %d %d\n", it->x, it->y, it->z);
  return 0;
}

### Prompt

In cpp, your task is to solve 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;
using ll = long long;
const int N = 2e5 + 5;
struct P {
  int x, y, z;
};
int n, mx, id, zy, yh, d[2][N], isd[N];
ll ans;
basic_string<int> G[N];
basic_string<P> res;
queue<int> q;
void dfs(int u, int fa, int d) {
  if (d > mx) mx = d, id = u;
  for (int v : G[u])
    if (v != fa) dfs(v, u, d + 1);
}
int mark(int u, int fa) {
  isd[u] = 1;
  if (u == yh) return 1;
  for (int v : G[u])
    if (v != fa && mark(v, u)) return 1;
  return isd[u] = 0;
}
void getd(int u, int fa, int f) {
  for (int v : G[u])
    if (v != fa) {
      if (f) {
        if (isd[v])
          res += {u, zy, u}, ans += d[0][u];
        else
          q.push(v);
      }
      d[f][v] = d[f][u] + 1, getd(v, u, f);
    }
}
int main() {
  scanf("%d", &n);
  for (int i = 1, a, b; i < n; ++i) scanf("%d%d", &a, &b), G[a] += b, G[b] += a;
  dfs(1, 0, 0), zy = id, mx = 0, dfs(id, 0, 0), yh = id;
  mark(zy, 0), getd(zy, 0, 0), getd(yh, 0, 1);
  reverse(res.begin(), res.end());
  while (!q.empty()) {
    int u = q.front();
    q.pop();
    if (d[0][u] > d[1][u])
      res += {u, zy, u}, ans += d[0][u];
    else
      res += {u, yh, u}, ans += d[1][u];
  }
  printf("%lld\n", ans);
  for (auto it = res.rbegin(); it != res.rend(); ++it)
    printf("%d %d %d\n", it->x, it->y, it->z);
  return 0;
}

```