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>
using namespace std;
int st, ed;
queue<int> q;
int edgenum, du[404040], yes[404040], jl[404040], tot, num, flag[404040], n,
    head[404040], vet[404040], jie[404040];
long long dis1[404040], dis2[404040], dis[404040];
long long ans = 0;
struct node {
  int u, v, w;
} a[404040];
void add(int u, int v) {
  edgenum++;
  vet[edgenum] = v;
  jie[edgenum] = head[u];
  head[u] = edgenum;
}
void dfs(int u, int ff, int dep) {
  int e = head[u];
  dis[u] = dep;
  while (e > 0) {
    int v = vet[e];
    if (v != ff) dfs(v, u, dep + 1);
    e = jie[e];
  }
}
void solve(int u, int ff) {
  int e = head[u];
  if (u == ed) {
    yes[u] = 1;
    num++;
    jl[num] = u;
    return;
  }
  while (e > 0) {
    int v = vet[e];
    if (v != ff) {
      solve(v, u);
      if (yes[v]) {
        yes[u] = 1;
        num++;
        jl[num] = u;
        return;
      }
    }
    e = jie[e];
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i < n; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    add(u, v);
    add(v, u);
    du[u]++;
    du[v]++;
  }
  dfs(1, 0, 0);
  st = 1;
  ed = 1;
  for (int i = 1; i <= n; i++)
    if (dis[i] > dis[st]) st = i;
  dfs(st, 0, 0);
  for (int i = 1; i <= n; i++)
    if (dis[i] > dis[ed]) ed = i;
  for (int i = 1; i <= n; i++) dis1[i] = dis[i];
  dfs(ed, 0, 0);
  for (int i = 1; i <= n; i++) dis2[i] = dis[i];
  for (int i = 1; i <= n; i++)
    if (du[i] == 1 && i != st && i != ed) {
      q.push(i);
    }
  while (!q.empty()) {
    int u = q.front();
    flag[u] = 1;
    q.pop();
    if (dis1[u] > dis2[u]) {
      ans += dis1[u];
      tot++;
      a[tot].u = u;
      a[tot].v = st;
      a[tot].w = u;
    } else {
      ans += dis2[u];
      tot++;
      a[tot].u = u;
      a[tot].v = ed;
      a[tot].w = u;
    }
    int e = head[u];
    while (e > 0) {
      int v = vet[e];
      if (flag[v] == 0) {
        du[v]--;
        if (du[v] == 1) q.push(v);
      }
      e = jie[e];
    }
  }
  solve(st, 0);
  int tmp = dis1[ed];
  for (int i = 1; i < num; i++) {
    ans += tmp;
    tmp--;
    tot++;
    a[tot].u = jl[i];
    a[tot].v = jl[num];
    a[tot].w = a[tot].u;
  }
  printf("%I64d\n", ans);
  for (int i = 1; i <= tot; i++) printf("%d %d %d\n", a[i].u, a[i].v, a[i].w);
  return 0;
}