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>
#pragma GCC optimize "trapv"
long long int power(long long int x, long long int b) {
  long long int p = 1;
  while (b > 0) {
    if (b & 1) {
      p = p * x;
      p %= 1000000007;
    }
    b >>= 1;
    x *= x;
    x %= 1000000007;
  }
  return p % 1000000007;
}
using namespace std;
struct ppp {
  char a;
  long long int b, c, d;
};
long long int N;
struct lex_compare {
  bool operator()(ppp p1, ppp p2) {
    if (p1.a == p2.a) {
      if (p1.b == p2.b)
        return p1.c < p2.c;
      else
        return p1.b < p2.b;
    } else {
      return p1.a < p2.a;
    }
  }
};
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  long long int n;
  cin >> n;
  long long int arr[n];
  for (long long int i = 0; i < n; i++) cin >> arr[i];
  long long int maxt = n;
  vector<vector<long long int> > ans;
  while (maxt--) {
    vector<vector<long long int> > blocks;
    int i = 0;
    while (i < n) {
      long long int s = arr[i];
      vector<long long int> temp;
      temp.push_back(s);
      i++;
      while (i < n && arr[i] == s + 1) {
        s = arr[i];
        i++;
        temp.push_back(s);
      }
      blocks.push_back(temp);
    }
    if (blocks.size() == 1) break;
    long long int minele = blocks[blocks.size() - 1].back();
    long long int next = minele + 1;
    if (next == n + 1) {
      vector<long long int> tempans;
      long long int ind = 0;
      for (int j = blocks.size() - 1; j >= 0; j--) {
        tempans.push_back(blocks[j].size());
        for (auto k : blocks[j]) {
          arr[ind++] = k;
        }
      }
      reverse(tempans.begin(), tempans.end());
      ans.push_back(tempans);
      continue;
    }
    long long int ind2nd = 0;
    while (ind2nd < blocks.size() && blocks[ind2nd][0] != next) {
      ind2nd++;
    }
    vector<long long int> tempans;
    long long int size1 = 0;
    i = 0;
    while (i < ind2nd) {
      size1 += blocks[i].size();
      i++;
    }
    if (size1 != 0) tempans.push_back(size1);
    size1 = 0;
    i = ind2nd;
    while (i < blocks.size() - 1) {
      size1 += blocks[i].size();
      i++;
    }
    if (size1 != 0) tempans.push_back(size1);
    tempans.push_back(blocks.back().size());
    ans.push_back(tempans);
    int ind = 0;
    for (auto j : blocks.back()) {
      arr[ind++] = j;
    }
    for (int j = ind2nd; j < blocks.size() - 1; j++) {
      for (auto k : blocks[j]) {
        arr[ind++] = k;
      }
    }
    for (int j = 0; j < ind2nd; j++) {
      for (auto k : blocks[j]) {
        arr[ind++] = k;
      }
    }
  }
  cout << ans.size() << "\n";
  for (auto i : ans) {
    cout << i.size() << " ";
    for (auto j : i) cout << j << " ";
    cout << "\n";
  }
  return 0;
}