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:
import sys

def minp():
	return sys.stdin.readline().strip()

def mint():
	return int(minp())

def mints():
	return map(int, minp().split())


def solve():
	n = mint()
	c = list(mints())
	if n == 1:
		print(0)
		return
	#if n == 2:
	#	if c != [1,2]:
	#		print(1)
	#		print(2,1,1)
	#	else:
	#		print(0)
	want = list(range(1,n+1)) if n % 2 == 0 else list(range(n,0,-1))
	zs = 1
	ze = n+1
	if {c.index(1), want.index(1)} == {0,n-1}:
		print(n-1)
		print(n, *([1]*n))
		c.reverse()
		zs = 2
		ze = n
	else:
		print(n)
	for i in range(zs,ze):
		idx = c.index(i)
		idxw = want.index(i)
		#print(c, want, idx, idxw)
		if idx == idxw:
			print(n, *([1]*n))
			c.reverse()
		else:
			if idx > idxw:
				idxw, idx = idx, idxw
			z = [1]*idx+[idxw-idx+1]+[1]*(n-idxw-1)
			#print(z)
			if sum(z) != n:
				raise Exception("wut")
			print(len(z), end=' ')
			print(*z)
			p = n
			cn = []
			for jj in range(len(z)-1,-1,-1):
				j = z[jj]
				for i in range(p-j, p):
					cn.append(c[i])
				p -= j
			c = cn
			#print('c',c)
		want.reverse()
	#print(c)

solve()