#include <bits/stdc++.h>
using namespace std;
const int N = 200010;
struct edge {
  int y, nex;
} s[N << 1];
int first[N], len = 0, n, mx, p;
int a[N], tot, t;
pair<int, int> ans[N];
long long op = 0;
void ins(int x, int y) {
  s[++len] = (edge){y, first[x]};
  first[x] = len;
}
void dfs(int x, int fa, int dep) {
  if (dep > mx) mx = dep, p = x;
  for (int i = first[x]; i != 0; i = s[i].nex)
    if (s[i].y != fa) dfs(s[i].y, x, dep + 1);
}
void ga(int x, int tar, int fa) {
  if (x == tar) {
    a[++tot] = tar;
    return;
  }
  for (int i = first[x]; i; i = s[i].nex)
    if (s[i].y != fa) {
      ga(s[i].y, tar, x);
      if (tot) {
        a[++tot] = x;
        return;
      }
    }
}
void gas(int x, int fa, int pos, int dep) {
  for (int i = first[x]; i; i = s[i].nex)
    if (s[i].y != fa) gas(s[i].y, x, pos, dep + 1);
  op += dep;
  ans[++t] = make_pair(pos, x);
}
int main() {
  scanf("%d", &n);
  int x, y;
  for (int i = 1; i < n; i++) scanf("%d %d", &x, &y), ins(x, y), ins(y, x);
  mx = 0;
  dfs(1, 0, 0);
  mx = 0;
  x = p;
  dfs(p, 0, 0);
  ga(x, p, 0);
  for (int i = 1; i <= tot; i++) {
    for (int j = first[a[i]]; j; j = s[j].nex)
      if (s[j].y != a[i - 1] && s[j].y != a[i + 1])
        gas(s[j].y, a[i], i <= tot / 2 ? a[tot] : a[1], max(i, tot - i + 1));
  }
  printf("%lld\n", op + 1ll * tot * (tot - 1) / 2);
  for (int i = 1; i <= t; i++)
    printf("%d %d %d\n", ans[i].first, ans[i].second, ans[i].second);
  for (int i = 1; i <= tot - 1; i++) printf("%d %d %d\n", a[tot], a[i], a[i]);
}

### 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;
const int N = 200010;
struct edge {
  int y, nex;
} s[N << 1];
int first[N], len = 0, n, mx, p;
int a[N], tot, t;
pair<int, int> ans[N];
long long op = 0;
void ins(int x, int y) {
  s[++len] = (edge){y, first[x]};
  first[x] = len;
}
void dfs(int x, int fa, int dep) {
  if (dep > mx) mx = dep, p = x;
  for (int i = first[x]; i != 0; i = s[i].nex)
    if (s[i].y != fa) dfs(s[i].y, x, dep + 1);
}
void ga(int x, int tar, int fa) {
  if (x == tar) {
    a[++tot] = tar;
    return;
  }
  for (int i = first[x]; i; i = s[i].nex)
    if (s[i].y != fa) {
      ga(s[i].y, tar, x);
      if (tot) {
        a[++tot] = x;
        return;
      }
    }
}
void gas(int x, int fa, int pos, int dep) {
  for (int i = first[x]; i; i = s[i].nex)
    if (s[i].y != fa) gas(s[i].y, x, pos, dep + 1);
  op += dep;
  ans[++t] = make_pair(pos, x);
}
int main() {
  scanf("%d", &n);
  int x, y;
  for (int i = 1; i < n; i++) scanf("%d %d", &x, &y), ins(x, y), ins(y, x);
  mx = 0;
  dfs(1, 0, 0);
  mx = 0;
  x = p;
  dfs(p, 0, 0);
  ga(x, p, 0);
  for (int i = 1; i <= tot; i++) {
    for (int j = first[a[i]]; j; j = s[j].nex)
      if (s[j].y != a[i - 1] && s[j].y != a[i + 1])
        gas(s[j].y, a[i], i <= tot / 2 ? a[tot] : a[1], max(i, tot - i + 1));
  }
  printf("%lld\n", op + 1ll * tot * (tot - 1) / 2);
  for (int i = 1; i <= t; i++)
    printf("%d %d %d\n", ans[i].first, ans[i].second, ans[i].second);
  for (int i = 1; i <= tot - 1; i++) printf("%d %d %d\n", a[tot], a[i], a[i]);
}

```