#include <bits/stdc++.h>
using namespace std;
int height[200005];
int lowest[200005];
int parent[200005];
int dist[200005][2];
int diameter = 0;
std::vector<int> G[200005];
int e1 = 1;
int e2 = 1;
void dfs(int u, int par = -1, int l = 0) {
  height[u] = l;
  lowest[u] = u;
  parent[u] = par;
  for (int i = 0; i < G[u].size(); i++) {
    if (G[u][i] == par) continue;
    dfs(G[u][i], u, l + 1);
    if (height[lowest[u]] < height[lowest[G[u][i]]]) {
      lowest[u] = lowest[G[u][i]];
    }
  }
}
void dfs2(int u, int par, int ai, int cdist = 0) {
  dist[u][ai] = cdist;
  for (int i = 0; i < G[u].size(); i++) {
    if (G[u][i] == par) continue;
    dfs2(G[u][i], u, ai, cdist + 1);
  }
}
std::vector<pair<pair<int, int>, int> > ans;
long long sum = 0;
int diaLength;
void dfs3(int u, int par = -1) {
  for (int i = 0; i < G[u].size(); i++) {
    if (G[u][i] == par) continue;
    dfs3(G[u][i], u);
  }
  if (dist[u][0] + dist[u][1] != diaLength) {
    sum += max(dist[u][0], dist[u][1]);
    if (dist[u][0] > dist[u][1])
      ans.push_back(pair<pair<int, int>, int>(pair<int, int>(e1, u), u));
    else
      ans.push_back(pair<pair<int, int>, int>(pair<int, int>(e2, u), u));
  }
}
int main() {
  int N;
  int u, v;
  scanf("%d", &N);
  for (int i = 0; i < N - 1; i++) {
    scanf("%d%d", &u, &v);
    G[u].push_back(v);
    G[v].push_back(u);
  }
  dfs(1);
  e1 = lowest[1];
  dfs(e1);
  e2 = lowest[e1];
  diaLength = height[e2];
  dfs2(e1, -1, 0);
  dfs2(e2, -1, 1);
  dfs3(e1);
  int cur = e2;
  while (cur != e1) {
    sum += dist[cur][0];
    ans.push_back(pair<pair<int, int>, int>(pair<int, int>(e1, cur), cur));
    cur = parent[cur];
  }
  printf("%lld\n", sum);
  for (int i = 0; i < ans.size(); i++) {
    printf("%d %d %d\n", ans[i].first.first, ans[i].first.second,
           ans[i].second);
  }
  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;
int height[200005];
int lowest[200005];
int parent[200005];
int dist[200005][2];
int diameter = 0;
std::vector<int> G[200005];
int e1 = 1;
int e2 = 1;
void dfs(int u, int par = -1, int l = 0) {
  height[u] = l;
  lowest[u] = u;
  parent[u] = par;
  for (int i = 0; i < G[u].size(); i++) {
    if (G[u][i] == par) continue;
    dfs(G[u][i], u, l + 1);
    if (height[lowest[u]] < height[lowest[G[u][i]]]) {
      lowest[u] = lowest[G[u][i]];
    }
  }
}
void dfs2(int u, int par, int ai, int cdist = 0) {
  dist[u][ai] = cdist;
  for (int i = 0; i < G[u].size(); i++) {
    if (G[u][i] == par) continue;
    dfs2(G[u][i], u, ai, cdist + 1);
  }
}
std::vector<pair<pair<int, int>, int> > ans;
long long sum = 0;
int diaLength;
void dfs3(int u, int par = -1) {
  for (int i = 0; i < G[u].size(); i++) {
    if (G[u][i] == par) continue;
    dfs3(G[u][i], u);
  }
  if (dist[u][0] + dist[u][1] != diaLength) {
    sum += max(dist[u][0], dist[u][1]);
    if (dist[u][0] > dist[u][1])
      ans.push_back(pair<pair<int, int>, int>(pair<int, int>(e1, u), u));
    else
      ans.push_back(pair<pair<int, int>, int>(pair<int, int>(e2, u), u));
  }
}
int main() {
  int N;
  int u, v;
  scanf("%d", &N);
  for (int i = 0; i < N - 1; i++) {
    scanf("%d%d", &u, &v);
    G[u].push_back(v);
    G[v].push_back(u);
  }
  dfs(1);
  e1 = lowest[1];
  dfs(e1);
  e2 = lowest[e1];
  diaLength = height[e2];
  dfs2(e1, -1, 0);
  dfs2(e2, -1, 1);
  dfs3(e1);
  int cur = e2;
  while (cur != e1) {
    sum += dist[cur][0];
    ans.push_back(pair<pair<int, int>, int>(pair<int, int>(e1, cur), cur));
    cur = parent[cur];
  }
  printf("%lld\n", sum);
  for (int i = 0; i < ans.size(); i++) {
    printf("%d %d %d\n", ans[i].first.first, ans[i].first.second,
           ans[i].second);
  }
  return 0;
}

```