#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 5;
vector<int> adj[N], brg;
vector<pair<int, int> > ANS;
int H[2][N], T[N], P[N], n, a, b, tmp, rt[2], k;
long long ans;
bool C(int u) { return (u && u != rt[0] && u != rt[1]); }
void rtfind(int u, int p, int d) {
  if (d >= tmp) rt[k] = u, tmp = d;
  for (auto x : adj[u])
    if (x != p) rtfind(x, u, d + 1);
}
void hset(int u, int p, int d) {
  H[k][u] = d;
  for (auto x : adj[u])
    if (x != p) hset(x, u, d + 1);
}
void DFS(int u, int p) {
  if (p != -1) T[u]++, P[u] = p;
  for (auto x : adj[u])
    if (x != p) DFS(x, u), T[u]++;
  if (C(u) && T[u] == 1) brg.push_back(u);
}
void Do() {
  while (brg.size()) {
    a = brg.back(), brg.pop_back();
    if (H[0][a] > H[1][a])
      ANS.push_back({rt[0], a});
    else
      ANS.push_back({rt[1], a});
    ans += max(H[0][a], H[1][a]);
    if (C(P[a]) && --T[P[a]] == 1) brg.push_back(P[a]);
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i < n; i++) {
    scanf("%d %d", &a, &b);
    adj[a].push_back(b), adj[b].push_back(a);
  }
  rtfind(1, -1, 0), k++, tmp = 0, rtfind(rt[0], -1, 0);
  k = 0, hset(rt[0], -1, 0), k++, hset(rt[1], -1, 0);
  DFS(rt[0], -1);
  Do();
  ans += H[0][rt[1]], ANS.push_back({rt[0], rt[1]});
  while (P[rt[1]] != rt[0]) {
    ans += H[0][P[rt[1]]], ANS.push_back({rt[0], P[rt[1]]});
    rt[1] = P[rt[1]];
  }
  printf("%I64d\n", ans);
  for (auto x : ANS) printf("%d %d %d\n", x.first, x.second, x.second);
}

### Prompt

Please formulate 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 = 2e5 + 5;
vector<int> adj[N], brg;
vector<pair<int, int> > ANS;
int H[2][N], T[N], P[N], n, a, b, tmp, rt[2], k;
long long ans;
bool C(int u) { return (u && u != rt[0] && u != rt[1]); }
void rtfind(int u, int p, int d) {
  if (d >= tmp) rt[k] = u, tmp = d;
  for (auto x : adj[u])
    if (x != p) rtfind(x, u, d + 1);
}
void hset(int u, int p, int d) {
  H[k][u] = d;
  for (auto x : adj[u])
    if (x != p) hset(x, u, d + 1);
}
void DFS(int u, int p) {
  if (p != -1) T[u]++, P[u] = p;
  for (auto x : adj[u])
    if (x != p) DFS(x, u), T[u]++;
  if (C(u) && T[u] == 1) brg.push_back(u);
}
void Do() {
  while (brg.size()) {
    a = brg.back(), brg.pop_back();
    if (H[0][a] > H[1][a])
      ANS.push_back({rt[0], a});
    else
      ANS.push_back({rt[1], a});
    ans += max(H[0][a], H[1][a]);
    if (C(P[a]) && --T[P[a]] == 1) brg.push_back(P[a]);
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i < n; i++) {
    scanf("%d %d", &a, &b);
    adj[a].push_back(b), adj[b].push_back(a);
  }
  rtfind(1, -1, 0), k++, tmp = 0, rtfind(rt[0], -1, 0);
  k = 0, hset(rt[0], -1, 0), k++, hset(rt[1], -1, 0);
  DFS(rt[0], -1);
  Do();
  ans += H[0][rt[1]], ANS.push_back({rt[0], rt[1]});
  while (P[rt[1]] != rt[0]) {
    ans += H[0][P[rt[1]]], ANS.push_back({rt[0], P[rt[1]]});
    rt[1] = P[rt[1]];
  }
  printf("%I64d\n", ans);
  for (auto x : ANS) printf("%d %d %d\n", x.first, x.second, x.second);
}

```