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:
from bisect import bisect_left as _bisect_left
import sys as _sys


def main():
    n, = _read_ints()
    c = tuple(_read_ints())
    operations = how_to_sort(c)
    print(len(operations))
    for operation in operations:
        print(len(operation), *operation)


def _read_line():
    result = _sys.stdin.readline()
    assert result[-1] == "\n"
    return result[:-1]


def _read_ints():
    return map(int, _read_line().split())


def how_to_sort(sequence):
    n = len(sequence)
    sequence = list(sequence)
    operations = []
    
    if n <= 1:
        return operations
    
    if n == 2:
        if sequence[0] > sequence[1]:
            operations.append((1, 1))
        return operations
    
    if n % 2 != 0:
        i_mid = n // 2
        left_part, sorted_part, right_part = \
            sequence[:i_mid], [sequence[i_mid]], sequence[i_mid+1:]
    else:
        assert n >= 4
        i_left_mid, i_right_mid = n//2 - 1, n//2
        if sequence[i_left_mid] > sequence[i_right_mid]:
            operations.append((n//2 - 1, 1, 1, n//2 - 1))
            sequence = (
                sequence[i_right_mid+1:]
                +
                [sequence[i_right_mid], sequence[i_left_mid]]
                +
                sequence[:i_left_mid]
            )
        left_part, sorted_part, right_part = \
            sequence[:i_left_mid], sequence[i_left_mid:i_right_mid+1], sequence[i_right_mid+1:]
    
    assert len(left_part) == len(right_part)
    
    while len(sorted_part) < n:
        sorted_n = len(sorted_part)
        lx = left_part.pop(-1)
        rx = right_part.pop(0)
        ilx = _bisect_left(sorted_part, lx)
        irx = _bisect_left(sorted_part, rx)
        if lx <= rx:
            operations.append(
                (
                    len(left_part),
                    1,
                    ilx,
                    len(sorted_part[ilx:irx]),
                    len(sorted_part[irx:]),
                    1,
                    len(right_part)
                )
            )
            operations.append(
                (
                    len(right_part),
                    len(sorted_part[irx:]) + 1,
                    len(sorted_part[ilx:irx]),
                    ilx + 1,
                    len(left_part)
                )
            )
        else:
            ilx, irx = irx, ilx
            operations.append(
                (
                    len(left_part),
                    ilx + 1,
                    len(sorted_part[ilx:irx]),
                    len(sorted_part[irx:]) + 1,
                    len(right_part)
                )
            )
            operations.append(
                (
                    len(right_part),
                    len(sorted_part[irx:]),
                    len(sorted_part[ilx:irx]) + 2,
                    ilx,
                    len(left_part)
                )
            )
        sorted_part.append(lx)
        sorted_part.append(rx)
        sorted_part.sort()
    
    for i, operation in enumerate(operations):
        operation = tuple(filter(lambda x: x != 0, operation))
        operations[i] = operation
    
    operations = list(filter(lambda operation: len(operation) > 1, operations))
    
    return operations


if __name__ == '__main__':
    main()