#include <bits/stdc++.h>
using namespace std;
int n, head[200005], cnt, st, ed, num[200005], tot, from[200005], f[200005],
    a[200005], b[200005], c[200005], dis[200005];
long long ans;
struct edge {
  int to, next;
} e[200005 << 1];
bool vis[200005];
template <typename T>
inline T read() {
  T x = 0, w = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') w = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
  return x * w;
}
inline void adde(int u, int v) { e[++cnt] = (edge){v, head[u]}, head[u] = cnt; }
void dfs(int u, int fa) {
  for (int v, i = head[u]; i; i = e[i].next) {
    v = e[i].to;
    if (v == fa) continue;
    from[v] = u, f[v] = f[u] + 1;
    dfs(v, u);
  }
}
void dfs1(int u, int fa) {
  for (int v, i = head[u]; i; i = e[i].next) {
    v = e[i].to;
    if (v == fa) continue;
    dis[v] = dis[u] + 1;
    dfs1(v, u);
  }
}
void dfs2(int u, int fa) {
  for (int v, i = head[u]; i; i = e[i].next)
    if ((v = e[i].to) != fa) dfs2(v, u);
  if (!vis[u]) {
    if (dis[u] > f[u])
      ans += dis[u], a[++cnt] = ed, b[cnt] = u, c[cnt] = u;
    else
      ans += f[u], a[++cnt] = st, b[cnt] = u, c[cnt] = u;
  }
}
int main() {
  n = read<int>();
  for (int u, v, i = 1; i < n; i++) {
    u = read<int>(), v = read<int>();
    adde(u, v), adde(v, u);
  }
  dfs(1, 0), st = 1;
  for (int i = 1; i <= n; i++)
    if (f[i] > f[st]) st = i;
  memset(f, 0, sizeof(f));
  dfs(st, 0), ed = st;
  for (int i = 1; i <= n; i++)
    if (f[i] > f[ed]) ed = i;
  for (int tmp = ed; tmp != st; tmp = from[tmp]) vis[num[++tot] = tmp] = 1;
  vis[num[++tot] = st] = 1;
  dfs1(ed, 0);
  cnt = 0, dfs2(st, 0);
  for (int i = tot; i >= 2; i--)
    ans += i - 1, a[++cnt] = num[1], b[cnt] = num[i], c[cnt] = num[i];
  printf("%lld\n", ans);
  for (int i = 1; i <= cnt; i++) printf("%d %d %d\n", a[i], b[i], c[i]);
  return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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 n, head[200005], cnt, st, ed, num[200005], tot, from[200005], f[200005],
    a[200005], b[200005], c[200005], dis[200005];
long long ans;
struct edge {
  int to, next;
} e[200005 << 1];
bool vis[200005];
template <typename T>
inline T read() {
  T x = 0, w = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') w = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
  return x * w;
}
inline void adde(int u, int v) { e[++cnt] = (edge){v, head[u]}, head[u] = cnt; }
void dfs(int u, int fa) {
  for (int v, i = head[u]; i; i = e[i].next) {
    v = e[i].to;
    if (v == fa) continue;
    from[v] = u, f[v] = f[u] + 1;
    dfs(v, u);
  }
}
void dfs1(int u, int fa) {
  for (int v, i = head[u]; i; i = e[i].next) {
    v = e[i].to;
    if (v == fa) continue;
    dis[v] = dis[u] + 1;
    dfs1(v, u);
  }
}
void dfs2(int u, int fa) {
  for (int v, i = head[u]; i; i = e[i].next)
    if ((v = e[i].to) != fa) dfs2(v, u);
  if (!vis[u]) {
    if (dis[u] > f[u])
      ans += dis[u], a[++cnt] = ed, b[cnt] = u, c[cnt] = u;
    else
      ans += f[u], a[++cnt] = st, b[cnt] = u, c[cnt] = u;
  }
}
int main() {
  n = read<int>();
  for (int u, v, i = 1; i < n; i++) {
    u = read<int>(), v = read<int>();
    adde(u, v), adde(v, u);
  }
  dfs(1, 0), st = 1;
  for (int i = 1; i <= n; i++)
    if (f[i] > f[st]) st = i;
  memset(f, 0, sizeof(f));
  dfs(st, 0), ed = st;
  for (int i = 1; i <= n; i++)
    if (f[i] > f[ed]) ed = i;
  for (int tmp = ed; tmp != st; tmp = from[tmp]) vis[num[++tot] = tmp] = 1;
  vis[num[++tot] = st] = 1;
  dfs1(ed, 0);
  cnt = 0, dfs2(st, 0);
  for (int i = tot; i >= 2; i--)
    ans += i - 1, a[++cnt] = num[1], b[cnt] = num[i], c[cnt] = num[i];
  printf("%lld\n", ans);
  for (int i = 1; i <= cnt; i++) printf("%d %d %d\n", a[i], b[i], c[i]);
  return 0;
}

```