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 PYTHON3):

Solution:
n = int(input())

c = [int(x) for x in input().split()]

ops = []
correct_side = c[0] == 1
touched = False

def make_selection(selection):
    global correct_side, c
    #print("Raw selection is", selection)
    selection = sorted(x for x in selection if x[0] <= x[1])
    if len(selection) == 1:
        #print("Selection is too small", selection)
        return
    
    ops.append(selection)
    #print(len(ops), selection)

    p = []
    for b, e in reversed(selection):
        p += c[b:e+1]

    c = p

    correct_side = c[0] == 1
    #print("correct_side is now", correct_side)
    #print(c)
    

for i in range(1, n + 1):
    # We must now find the ith card and move it to the correct
    # spot.

    selection = []

    # Make sure solved stuff stays solved
    for j in range(1, i):
        if correct_side:
            selection.append((j-1, j-1))
        else:
            selection.append((n-j, n-j))

    #print("Protect with", selection)

    # Find the card we want
    found = c.index(i)

    #print("Target", i, "is at index", found)

    # Do we need to do an op?
    if (correct_side and found == i - 1) or (not correct_side and found == n-i):
        continue

    #print("Executing... correct_side? ", correct_side)

    # Ok, we do.
    if correct_side:
        selection.append((i - 1, found))
        selection.append((found + 1, n - 1))
    else:
        selection.append((0, found - 1))
        selection.append((found, n-i))

    make_selection(selection)
            
if not correct_side:
    make_selection([(x,x) for x in range(0, n)])

print(len(ops))
for selection in ops:
    print(len(selection), end=' ')
    for b, e in selection:
        print(e - b + 1, end=' ')
    print()