#include <bits/stdc++.h>
using namespace std;
const int N = 200005;
int n, a, b, dA[N], dB[N], deg[N];
vector<int> g[N];
int mx = -1, mxNode = -1;
void dfs(int node, int p, int dist, int d[]) {
  if (dist > mx) mx = dist, mxNode = node;
  d[node] = dist;
  for (int ch : g[node])
    if (ch != p) dfs(ch, node, dist + 1, d);
}
struct node {
  int a, b, c;
};
vector<node> ans;
int main() {
  scanf("%d", &n);
  for (int(i) = (2); (i) <= (int)(n); (i)++) {
    int a, b;
    scanf("%d%d", &a, &b);
    g[a].push_back(b), g[b].push_back(a);
    ++deg[a], ++deg[b];
  }
  mx = -1, mxNode = -1;
  dfs(1, -1, 0, dA);
  a = mxNode;
  mx = -1, mxNode = -1;
  dfs(a, -1, 0, dA);
  b = mxNode;
  dfs(b, -1, 0, dB);
  queue<int> q;
  for (int(i) = (1); (i) <= (int)(n); (i)++)
    if (i != a && i != b && deg[i] == 1) q.push(i);
  long long ansSum = 0;
  while (!q.empty()) {
    int cur = q.front();
    q.pop();
    if (dA[cur] > dB[cur])
      ans.push_back({cur, a, cur}), ansSum += dA[cur];
    else
      ans.push_back({cur, b, cur}), ansSum += dB[cur];
    for (int ch : g[cur])
      if (--deg[ch] == 1 && ch != a && ch != b) q.push(ch);
  }
  q.push(b);
  while (!q.empty()) {
    int cur = q.front();
    q.pop();
    ansSum += dA[cur];
    ans.push_back({cur, a, cur});
    for (int ch : g[cur])
      if (--deg[ch] == 1 && ch != a) q.push(ch);
  }
  printf("%lld\n", ansSum);
  for (node x : ans) printf("%d %d %d\n", x.a, x.b, x.c);
  return 0;
}

### Prompt

Your challenge is to write 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 N = 200005;
int n, a, b, dA[N], dB[N], deg[N];
vector<int> g[N];
int mx = -1, mxNode = -1;
void dfs(int node, int p, int dist, int d[]) {
  if (dist > mx) mx = dist, mxNode = node;
  d[node] = dist;
  for (int ch : g[node])
    if (ch != p) dfs(ch, node, dist + 1, d);
}
struct node {
  int a, b, c;
};
vector<node> ans;
int main() {
  scanf("%d", &n);
  for (int(i) = (2); (i) <= (int)(n); (i)++) {
    int a, b;
    scanf("%d%d", &a, &b);
    g[a].push_back(b), g[b].push_back(a);
    ++deg[a], ++deg[b];
  }
  mx = -1, mxNode = -1;
  dfs(1, -1, 0, dA);
  a = mxNode;
  mx = -1, mxNode = -1;
  dfs(a, -1, 0, dA);
  b = mxNode;
  dfs(b, -1, 0, dB);
  queue<int> q;
  for (int(i) = (1); (i) <= (int)(n); (i)++)
    if (i != a && i != b && deg[i] == 1) q.push(i);
  long long ansSum = 0;
  while (!q.empty()) {
    int cur = q.front();
    q.pop();
    if (dA[cur] > dB[cur])
      ans.push_back({cur, a, cur}), ansSum += dA[cur];
    else
      ans.push_back({cur, b, cur}), ansSum += dB[cur];
    for (int ch : g[cur])
      if (--deg[ch] == 1 && ch != a && ch != b) q.push(ch);
  }
  q.push(b);
  while (!q.empty()) {
    int cur = q.front();
    q.pop();
    ansSum += dA[cur];
    ans.push_back({cur, a, cur});
    for (int ch : g[cur])
      if (--deg[ch] == 1 && ch != a) q.push(ch);
  }
  printf("%lld\n", ansSum);
  for (node x : ans) printf("%d %d %d\n", x.a, x.b, x.c);
  return 0;
}

```