#include <bits/stdc++.h>
using namespace std;
const long long maxn = 2e5 + 5;
long long p[maxn];
long long dis[maxn];
long long dep[maxn];
long long par[maxn];
bool vis[maxn];
long long res[maxn];
long long n, cnt, maxd, rt1, rt2;
long long ans;
long long l;
struct node {
  long long v, next;
} e[maxn * 2];
void addEdge(long long u, long long v) {
  cnt++;
  e[cnt].v = v;
  e[cnt].next = p[u];
  p[u] = cnt;
}
void dfs(long long u, long long f) {
  for (long long i = p[u]; i != -1; i = e[i].next) {
    if (e[i].v == f) continue;
    par[e[i].v] = u;
    dep[e[i].v] = dep[u] + 1;
    if (dep[e[i].v] > dep[maxd]) maxd = e[i].v;
    dfs(e[i].v, u);
  }
}
void dfs2(long long u, long long f) {
  for (long long i = p[u]; i != -1; i = e[i].next) {
    if (e[i].v == f) continue;
    dis[e[i].v] = dis[u] + 1;
    dfs2(e[i].v, u);
  }
}
void dfs3(long long u, long long f) {
  for (long long i = p[u]; i != -1; i = e[i].next) {
    if (e[i].v == f) continue;
    dfs3(e[i].v, u);
  }
  if (!vis[u]) cout << u << ' ' << res[u] << ' ' << u << '\n';
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  memset(p, -1, sizeof(p));
  cin >> n;
  for (long long i = 1; i < n; i++) {
    long long u, v;
    cin >> u >> v;
    addEdge(u, v);
    addEdge(v, u);
  }
  dfs(1, -1);
  dep[maxd] = 0;
  rt1 = maxd;
  par[maxd] = -1;
  dfs(maxd, -1);
  rt2 = maxd;
  l = dep[rt2];
  for (long long i = rt2; i != -1; i = par[i]) vis[i] = 1;
  dfs2(rt2, -1);
  for (long long i = 1; i <= n; i++) {
    if (vis[i] == 0) {
      if (dep[i] > dis[i]) {
        res[i] = rt1;
        ans += dep[i];
      } else {
        res[i] = rt2;
        ans += dis[i];
      }
    }
  }
  ans += (l + 1) * l / 2;
  cout << ans << '\n';
  dfs3(rt1, -1);
  for (long long i = rt2; i != rt1; i = par[i])
    cout << i << ' ' << rt1 << ' ' << i << '\n';
}

### 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 long long maxn = 2e5 + 5;
long long p[maxn];
long long dis[maxn];
long long dep[maxn];
long long par[maxn];
bool vis[maxn];
long long res[maxn];
long long n, cnt, maxd, rt1, rt2;
long long ans;
long long l;
struct node {
  long long v, next;
} e[maxn * 2];
void addEdge(long long u, long long v) {
  cnt++;
  e[cnt].v = v;
  e[cnt].next = p[u];
  p[u] = cnt;
}
void dfs(long long u, long long f) {
  for (long long i = p[u]; i != -1; i = e[i].next) {
    if (e[i].v == f) continue;
    par[e[i].v] = u;
    dep[e[i].v] = dep[u] + 1;
    if (dep[e[i].v] > dep[maxd]) maxd = e[i].v;
    dfs(e[i].v, u);
  }
}
void dfs2(long long u, long long f) {
  for (long long i = p[u]; i != -1; i = e[i].next) {
    if (e[i].v == f) continue;
    dis[e[i].v] = dis[u] + 1;
    dfs2(e[i].v, u);
  }
}
void dfs3(long long u, long long f) {
  for (long long i = p[u]; i != -1; i = e[i].next) {
    if (e[i].v == f) continue;
    dfs3(e[i].v, u);
  }
  if (!vis[u]) cout << u << ' ' << res[u] << ' ' << u << '\n';
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  memset(p, -1, sizeof(p));
  cin >> n;
  for (long long i = 1; i < n; i++) {
    long long u, v;
    cin >> u >> v;
    addEdge(u, v);
    addEdge(v, u);
  }
  dfs(1, -1);
  dep[maxd] = 0;
  rt1 = maxd;
  par[maxd] = -1;
  dfs(maxd, -1);
  rt2 = maxd;
  l = dep[rt2];
  for (long long i = rt2; i != -1; i = par[i]) vis[i] = 1;
  dfs2(rt2, -1);
  for (long long i = 1; i <= n; i++) {
    if (vis[i] == 0) {
      if (dep[i] > dis[i]) {
        res[i] = rt1;
        ans += dep[i];
      } else {
        res[i] = rt2;
        ans += dis[i];
      }
    }
  }
  ans += (l + 1) * l / 2;
  cout << ans << '\n';
  dfs3(rt1, -1);
  for (long long i = rt2; i != rt1; i = par[i])
    cout << i << ' ' << rt1 << ' ' << i << '\n';
}

```