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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
inline int read() {
  char c = getchar();
  int s = 0;
  while (!isdigit(c)) c = getchar();
  while (isdigit(c)) s = 10 * s + c - '0', c = getchar();
  return s;
}
const int maxn = 2e5 + 5;
int n, head[maxn], to[maxn << 1], nxt[maxn << 1], cnt;
void add(int u, int v) {
  to[++cnt] = v;
  nxt[cnt] = head[u];
  head[u] = cnt;
  return;
}
int maxdist = 0, d1, d2;
int dis[2][maxn];
void dfs1(int u, int fa, int dist, int &d, int tag) {
  if (dist > maxdist) d = u, maxdist = dist;
  if (tag > -1) dis[tag][u] = dist;
  for (int i = head[u]; i; i = nxt[i]) {
    int v = to[i];
    if (v == fa) continue;
    dfs1(v, u, dist + 1, d, tag);
  }
  return;
}
int diameter[maxn], tot = 0;
long long ret = 0;
struct node {
  int a, b, c;
  node() {}
  node(int u, int v, int w) { a = u, b = v, c = w; }
} ans[maxn];
bool dfs2(int u, int fa) {
  bool flag = 0;
  if (u == d2) return diameter[++cnt] = u, 1;
  for (int i = head[u]; i; i = nxt[i]) {
    int v = to[i];
    if (v == fa) continue;
    if (dfs2(v, u)) diameter[++cnt] = u, flag = 1;
  }
  if (!flag)
    if (dis[0][u] > dis[1][u])
      ans[++tot] = node(u, d1, u), ret += dis[0][u];
    else
      ans[++tot] = node(u, d2, u), ret += dis[1][u];
  return flag;
}
int main() {
  n = read();
  for (int i = 1; i <= n - 1; ++i) {
    int u = read(), v = read();
    add(u, v);
    add(v, u);
  }
  dfs1(1, 0, 0, d1, -1);
  maxdist = 0;
  dfs1(d1, 0, 0, d2, 0);
  dfs1(d2, 0, 0, d1, 1);
  cnt = 0;
  dfs2(d1, 0);
  for (int i = 1; i <= cnt - 1; ++i)
    ans[++tot] = node(diameter[i], d1, diameter[i]), ret += dis[0][diameter[i]];
  printf("%lld\n", ret);
  for (int i = 1; i <= tot; ++i)
    printf("%d %d %d\n", ans[i].a, ans[i].b, ans[i].c);
  return 0;
}