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;
bool ordenada(vector<int>& c) {
  for (int i = 1; i < c.size(); i++) {
    if (c[i] != c[i - 1] + 1) return false;
  }
  return true;
}
vector<int> nuevo(vector<pair<int, int> >& intervs, vector<int> c) {
  vector<int> res;
  for (int i = intervs.size() - 1; i >= 0; i--) {
    for (int j = intervs[i].first; j <= intervs[i].second; j++) {
      res.push_back(c[j]);
    }
  }
  return res;
}
int main() {
  int n;
  cin >> n;
  vector<int> c(n);
  for (int i = 0; i < n; i++) cin >> c[i];
  vector<vector<pair<int, int> > > ops;
  int ble = 0;
  while (!ordenada(c)) {
    for (int i = 1; i <= n; i++) {
      vector<pair<int, int> > intervs;
      int posi = -1;
      int posim1 = -1;
      for (int j = 0; j < n; j++) {
        if (c[j] == i) {
          posi = j;
        }
        if (c[j] == i + 1) {
          posim1 = j;
        }
      }
      if (posi > posim1) {
        if (posim1 != 0) intervs.push_back(make_pair(0, max(0, posim1 - 1)));
        int j = posim1 + 1;
        while (j < n && c[j] == c[j - 1] + 1) j++;
        intervs.push_back(make_pair(posim1, j - 1));
        intervs.push_back(make_pair(j, posi));
        if (posi != n - 1)
          intervs.push_back(make_pair(min(posi + 1, n - 1), n - 1));
        ops.push_back(intervs);
        c = nuevo(intervs, c);
        break;
      }
    }
  }
  cout << ops.size() << '\n';
  for (int i = 0; i < ops.size(); i++) {
    cout << ops[i].size();
    for (int j = 0; j < ops[i].size(); j++) {
      cout << ' ' << ops[i][j].second - ops[i][j].first + 1;
    }
    cout << '\n';
  }
}