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;
const long long mod = 1000000007;
const long long INF = 0x3f3f3f3f3f3f3f3f;
const double eps = 1e-6;
template <class T>
T gcd(T a, T b) {
  if (!b) return a;
  return gcd(b, a % b);
}
const int maxn = 2e5 + 10;
struct pnt {
  int x, y, del;
  pnt(int xx = 0, int yy = 0, int dd = 0) {
    x = xx;
    y = yy;
    del = dd;
  }
};
vector<pnt> vp;
int n, p[2], mx, on[maxn], d[maxn], fa[maxn];
long long ans = 0;
vector<int> edg[maxn];
void dfs1(int u, int f, int dis, int pos) {
  if (dis > mx) {
    p[pos] = u;
    mx = dis;
  }
  d[u] = dis;
  fa[u] = f;
  for (int i = 0; i < edg[u].size(); i++) {
    int v = edg[u][i];
    if (v == f) continue;
    dfs1(v, u, dis + 1, pos);
  }
}
void dfs2(int u, int f, int d1, int d2) {
  for (int i = 0; i < edg[u].size(); i++) {
    int v = edg[u][i];
    if (v == f || on[v]) continue;
    dfs2(v, u, d1 + 1, d2 + 1);
  }
  if (on[u]) {
    if (fa[u] != u) dfs2(fa[u], u, d1 + 1, d2 - 1);
    ans += d1;
    if (u != p[1]) vp.push_back(pnt(p[1], u, u));
    return;
  }
  if (d1 > d2) {
    ans += d1;
    vp.push_back(pnt(p[1], u, u));
  } else {
    ans += d2;
    vp.push_back(pnt(p[0], u, u));
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i < n; i++) {
    int u, v;
    scanf("%d", &u);
    scanf("%d", &v);
    edg[u].push_back(v);
    edg[v].push_back(u);
  }
  mx = -1;
  dfs1(1, 1, 0, 0);
  mx = -1;
  dfs1(p[0], p[0], 0, 1);
  on[p[0]] = 1;
  for (int i = p[1]; i != p[0]; i = fa[i]) {
    on[i] = 1;
  }
  dfs2(p[1], p[1], 0, d[p[1]]);
  printf("%I64d\n", ans);
  for (int i = 0; i < vp.size(); i++) {
    printf("%d %d %d\n", vp[i].x, vp[i].y, vp[i].del);
  }
  return 0;
}