Problem: You are given a deck of n cards numbered from 1 to n (not necessarily in this order in the deck). You have to sort the deck by repeating the following operation. 

  * Choose 2 ≤ k ≤ n and split the deck in k nonempty contiguous parts D_1, D_2,..., D_k (D_1 contains the first |D_1| cards of the deck, D_2 contains the following |D_2| cards and so on). Then reverse the order of the parts, transforming the deck into D_k, D_{k-1}, ..., D_2, D_1 (so, the first |D_k| cards of the new deck are D_k, the following |D_{k-1}| cards are D_{k-1} and so on). The internal order of each packet of cards D_i is unchanged by the operation. 



You have to obtain a sorted deck (i.e., a deck where the first card is 1, the second is 2 and so on) performing at most n operations. It can be proven that it is always possible to sort the deck performing at most n operations.

Examples of operation: The following are three examples of valid operations (on three decks with different sizes). 

  * If the deck is [3 6 2 1 4 5 7] (so 3 is the first card and 7 is the last card), we may apply the operation with k=4 and D_1=[3 6], D_2=[2 1 4], D_3=[5], D_4=[7]. Doing so, the deck becomes [7 5 2 1 4 3 6]. 
  * If the deck is [3 1 2], we may apply the operation with k=3 and D_1=[3], D_2=[1], D_3=[2]. Doing so, the deck becomes [2 1 3]. 
  * If the deck is [5 1 2 4 3 6], we may apply the operation with k=2 and D_1=[5 1], D_2=[2 4 3 6]. Doing so, the deck becomes [2 4 3 6 5 1]. 

Input

The first line of the input contains one integer n (1≤ n≤ 52) — the number of cards in the deck.

The second line contains n integers c_1, c_2, ..., c_n — the cards in the deck. The first card is c_1, the second is c_2 and so on.

It is guaranteed that for all i=1,...,n there is exactly one j∈\{1,...,n\} such that c_j = i.

Output

On the first line, print the number q of operations you perform (it must hold 0≤ q≤ n).

Then, print q lines, each describing one operation.

To describe an operation, print on a single line the number k of parts you are going to split the deck in, followed by the size of the k parts: |D_1|, |D_2|, ... , |D_k|. 

It must hold 2≤ k≤ n, and |D_i|≥ 1 for all i=1,...,k, and |D_1|+|D_2|+⋅⋅⋅ + |D_k| = n.

It can be proven that it is always possible to sort the deck performing at most n operations. If there are several ways to sort the deck you can output any of them.

Examples

Input


4
3 1 2 4


Output


2
3 1 2 1
2 1 3


Input


6
6 5 4 3 2 1


Output


1
6 1 1 1 1 1 1


Input


1
1


Output


0

Note

Explanation of the first testcase: Initially the deck is [3 1 2 4]. 

  * The first operation splits the deck as [(3) (1 2) (4)] and then transforms it into [4 1 2 3]. 
  * The second operation splits the deck as [(4) (1 2 3)] and then transforms it into [1 2 3 4]. 

Explanation of the second testcase: Initially the deck is [6 5 4 3 2 1]. 

  * The first (and only) operation splits the deck as [(6) (5) (4) (3) (2) (1)] and then transforms it into [1 2 3 4 5 6]. 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int oo = 1000000000;
const int N = 52 + 10;
const int M = oo + 7;
int n, v[N];
vector<vector<int> > an;
bool g;
int id(int x) {
  for (int i = 1; i <= n; ++i)
    if (v[i] == x) return i;
  return 0;
}
void upd() {
  assert(an.size());
  vector<int> &ls = an.back();
  assert(ls.size() > 1);
  int summ = 0;
  for (int i = 0; i < ls.size(); ++i) {
    summ += ls[i];
    if (!ls[i]) ls.erase(ls.begin() + i--);
  }
  assert(summ == n);
  vector<int> d[ls.size()];
  int j = 0, all = 0;
  for (int i = 1; i <= n; ++i) {
    if (i > all + ls[j]) all += ls[j++];
    d[j].push_back(v[i]);
  }
  j = 1;
  for (int i = ls.size() - 1; i >= 0; --i)
    for (int x : d[i]) v[j++] = x;
}
void go(vector<int> x) {
  an.push_back(x);
  upd();
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; ++i) cin >> v[i];
  if (n == 1) return cout << 0 << endl, 0;
  g = n & 1;
  int d = id(1);
  if (g && d > 1)
    go({d - 1, n - (d - 1)});
  else if (!g && d < n)
    go({d, n - d});
  int sz = 1;
  for (int l = 2, r = n; r - l;) {
    int cc = g ? r-- : l++;
    int e = id(cc);
    vector<int> cur;
    if (g)
      cur = {sz, e - sz, n - e};
    else
      cur = {e - 1, n - sz - (e - 1), sz};
    go(cur);
    g ^= 1;
    ++sz;
  }
  d = id(1);
  if (d > 1) go({d - 1, n - (d - 1)});
  cout << an.size() << endl;
  for (int i = 0; i < an.size(); ++i) {
    printf("%d", an[i].size());
    for (int x : an[i]) printf(" %d", x);
    puts("");
  }
  return 0;
}