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;
class Solution {
 public:
  vector<int> id, arr;
  int n;
  void flip(int left) {
    int right = n - left;
    for (int i = 0; i < n; i++) {
      if (id[i] < left)
        id[i] += right;
      else
        id[i] -= left;
    }
  }
  void flip(int c1, int c2) {
    if (c1 + c2 == n) {
      flip(c1);
      return;
    }
    int c3 = n - c1 - c2;
    for (int i = 0; i < n; i++) {
      if (id[i] < c1)
        id[i] += c2 + c3;
      else if (id[i] < c1 + c2)
        id[i] += -c1 + c3;
      else
        id[i] -= c1 + c2;
    }
  }
  void solve(int num, std::istream& in, std::ostream& out) {
    in >> n;
    arr.resize(n), id.resize(n);
    bool was = true;
    for (int i = 0; i < n; i++) {
      in >> arr[i];
      if (i && arr[i] < arr[i - 1]) was = false;
      id[arr[i] - 1] = i;
    }
    if (was) {
      out << 0;
      return;
    }
    int mid = n / 2;
    int gr = 1;
    int d = 0;
    int lo = mid;
    int hi = mid;
    vector<vector<int>> log;
    if (id[mid] != 0) {
      int cnt = id[mid];
      log.push_back({2, cnt, n - cnt});
      flip(cnt);
    }
    while (gr < n) {
      if (d == 0) {
        lo--;
        if (id[lo] == n - 1) {
          log.push_back({2, gr, n - gr});
          flip(gr);
        } else {
          log.push_back({3, gr, 1 + id[lo] - gr, n - 1 - id[lo]});
          flip(gr, 1 + id[lo] - gr);
        }
      } else {
        hi++;
        if (id[hi] == 0) {
          log.push_back({2, n - gr, gr});
          flip(n - gr);
        } else {
          log.push_back({3, id[hi], n - gr - id[hi], gr});
          flip(id[hi], n - gr - id[hi]);
        }
      }
      gr++;
      d ^= 1;
    }
    out << log.size() << endl;
    for (auto& vt : log) {
      for (int v : vt) {
        out << v << " ";
      }
      out << endl;
    }
  }
};
void solve(std::istream& in, std::ostream& out) {
  out << std::setprecision(12);
  Solution solution;
  solution.solve(0, in, out);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  istream& in = cin;
  ostream& out = cout;
  solve(in, out);
  return 0;
}