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]. 
#   Author: yumtam
#   Created at: 2020-10-11 00:47

from __future__ import division, print_function
_interactive = False

def main():
    n = int(input())
    ar = input_as_list()

    ans = []

    def op(lst):
        res = []
        i = 0
        for x in lst:
            res.append(ar[i:i+x])
            i += x
        ans.append(lst)
        return list(sum(reversed(res), []))


    back, front = 0, 1
    x = 1
    to = n%2
    while x <= n:
        i = ar.index(x)
        if to == back:
            ar = op([1]*(x-1) + [i+1-(x-1)] + [n-(i+1)])
        elif to == front:
            ar = op([i] + [n-i-(x-1)] + [1]*(x-1))
        x += 1
        to = 1-to

    for lst in ans:
        while 0 in lst:
            lst.remove(0)

    pans = []
    for lst in ans:
        if len(lst) > 1:
            pans += [lst]

    print(len(pans))
    for lst in pans:
        print(len(lst), end=' ')
        print(*lst)

    debug_print(ar)

# Constants
INF = float('inf')
MOD = 10**9+7

# Python3 equivalent names
import os, sys, itertools
if sys.version_info[0] < 3:
    input = raw_input
    range = xrange

    filter = itertools.ifilter
    map = itertools.imap
    zip = itertools.izip

# print-flush in interactive problems
if _interactive:
    flush = sys.stdout.flush
    def printf(*args, **kwargs):
        print(*args, **kwargs)
        flush()

# Debug print, only works on local machine
LOCAL = "LOCAL_" in os.environ
debug_print = (print) if LOCAL else (lambda *x, **y: None)

# Fast IO
if (not LOCAL) and (not _interactive):
    from io import BytesIO
    from atexit import register
    sys.stdin = BytesIO(os.read(0, os.fstat(0).st_size))
    sys.stdout = BytesIO()
    register(lambda: os.write(1, sys.stdout.getvalue()))
    input = lambda: sys.stdin.readline().rstrip('\r\n')

# Some utility functions(Input, N-dimensional lists, ...)
def input_as_list():
    return [int(x) for x in input().split()]

def input_with_offset(o):
    return [int(x)+o for x in input().split()]

def input_as_matrix(n, m):
    return [input_as_list() for _ in range(n)]

def array_of(f, *dim):
    return [array_of(f, *dim[1:]) for _ in range(dim[0])] if dim else f()

# Start of external code templates...
# End of external code templates.

main()
