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;
using namespace std;
const int maxn = 1000006;
struct edge {
  int v, nxt, val;
} e[maxn];
int cnt;
int head[maxn];
void add_edge(int u, int v, int w) {
  e[cnt].val = w;
  e[cnt].v = v;
  e[cnt].nxt = head[u];
  head[u] = cnt++;
}
int dis[maxn];
int pre[maxn];
int dfs(int x, int fa) {
  queue<int> que;
  que.push(x);
  memset(dis, -1, sizeof dis);
  dis[x] = 0;
  pre[x] = -1;
  int now = x;
  while (!que.empty()) {
    int u = que.front();
    que.pop();
    for (int i = head[u]; ~i; i = e[i].nxt) {
      int v = e[i].v;
      if (dis[v] == -1) {
        que.push(v);
        dis[v] = dis[u] + 1;
        pre[v] = u;
        if (dis[v] > dis[now]) {
          now = v;
        }
      }
    }
  }
  return now;
}
vector<int> link;
map<int, int> mp;
pair<int, int> pii[maxn];
void dfs1(int u, int fa, int aim) {
  queue<int> que;
  que.push(u);
  memset(dis, -1, sizeof dis);
  dis[u] = 0;
  while (!que.empty()) {
    int u = que.front();
    que.pop();
    for (int i = head[u]; ~i; i = e[i].nxt) {
      int v = e[i].v;
      if (dis[v] == -1) {
        que.push(v);
        dis[v] = dis[u] + 1;
        if (mp[v] == 0) {
          if (pii[v].first < dis[v]) pii[v] = {dis[v], aim};
        }
      }
    }
  }
}
long long ANS = 0;
vector<pair<int, int> > ans;
void DFS(int u, int fa) {
  for (int i = head[u]; ~i; i = e[i].nxt) {
    int v = e[i].v;
    if (mp[v]) continue;
    if (v == fa) continue;
    DFS(v, u);
    ANS += pii[v].first;
    ans.push_back({v, pii[v].second});
  }
}
int main() {
  cnt = 0;
  int n;
  memset(pre, 0, sizeof pre);
  scanf("%d", &n);
  memset(head, -1, sizeof head);
  for (int i = 1; i <= n - 1; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    add_edge(a, b, 1);
    add_edge(b, a, 1);
  }
  int a = dfs(1, -1);
  int b = dfs(a, -1);
  int tmp = b;
  vector<int> link;
  while (tmp != -1) {
    if (tmp != a) {
      pii[tmp] = {dis[tmp], a};
      link.push_back(tmp);
      ANS += dis[tmp];
    }
    mp[tmp] = 1;
    tmp = pre[tmp];
  }
  dfs1(a, -1, a);
  dfs1(b, -1, b);
  for (int i = 0; i < link.size(); i++) {
    DFS(link[i], -1);
  }
  for (int i = 0; i < link.size(); i++) {
    ans.push_back({link[i], pii[link[i]].second});
  }
  tmp = b;
  cout << ANS << endl;
  for (int i = 0; i < ans.size(); i++) {
    printf("%d %d %d\n", ans[i].first, ans[i].second, ans[i].first);
  }
}