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())
deck = [int(i) for i in input().split()]
ops = []
# for _ in range(n):
while True:
    # print(deck)
    grouping_starts = [0]  # consecutives
    for i in range(1, len(deck)):
        if deck[i] != deck[i-1] + 1:
            grouping_starts.append(i)
    if len(grouping_starts) == 1:
        # sorted
        break
    # always increase length of consecutive runs by putting first out-of-order run after
    # correct preceding run
    grouping_starts.append(n)
    if deck[0] != 1:
        for i in range(2, len(grouping_starts)):
            start = grouping_starts[i]
            if deck[start-1] == deck[0]-1:
                grouping_starts = [0, grouping_starts[i-1], grouping_starts[i], n]
                if grouping_starts[-2] == n:
                    grouping_starts.pop()
                break
    else:
        second_group_start_val = deck[grouping_starts[1]]
        for i in range(3, len(grouping_starts)):  # guaranteed to be length>2
            start = grouping_starts[i]
            if deck[start-1] == second_group_start_val-1:
                grouping_starts = [0, grouping_starts[1], grouping_starts[i-1], grouping_starts[i], n]
                if grouping_starts[-2] == n:
                    grouping_starts.pop()
                break
    op = [len(grouping_starts)-1]
    new_deck = []
    for i in range(len(grouping_starts)-1):
        start = grouping_starts[i]
        end = grouping_starts[i+1]
        # new_deck.extend(reversed(deck[start:end]))
        new_deck = deck[start:end] + new_deck
        op.append(end-start)
    # deck = list(reversed(new_deck))
    deck = new_deck
    ops.append(' '.join(str(i) for i in op))

assert len(ops) <= n
print(len(ops))
if ops:
    print(*ops, sep='\n')