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]. 
pocet = int(input())
d = list(map(int, input().split(" ")))
if pocet == 1:
	print('0')
else: 
	rev = False
	n = 0
	res = []	
	for i in range(pocet-1):
		#print(d)
		ind = d.index(i+1)
		ind2 = d.index(i+2)
		if ind != (ind2-1):
			n += 1
			if ind > ind2:
				doc = []
				if ind2 > 0:
					doc.append(str(ind2))
				if rev:
					doc.append(str(ind-ind2))
					doc.append(str(1))
					doc.append(str(i))
					if (ind+i) < (pocet - 1):
						doc.append(str(pocet-1-ind-i))
				else:
					doc.append(str(ind-i-ind2))
					doc.append(str(i+1))
					if ind < (pocet - 1):
						doc.append(str(pocet-1-ind))
				res.append(doc)
				rev = False
				#print(res)
			else:
				doc = []
				if rev:
					if ind > 0:
						doc.append(str(ind))
					doc.append(str(1))
					doc.append(str(i))
					doc.append(str(ind2-ind-i))
				else:
					if ind > i :
						doc.append(str(ind-i))
					doc.append(str(i+ind2-ind))
					doc.append(str(1))
				if ind2 < (pocet - 1):
					doc.append(str(pocet-1-ind2))
				res.append(doc)
				rev = True
				#print(res)
			docd = []
			inde = 0
			for j in reversed(range(len(doc))):
				for k in range(int(doc[j])):
					docd.append(d[len(d)-int(doc[j])+k-inde])
				inde += int(doc[j])
			d = docd
			#print(d)
	if d.index(pocet) == 0:
		res.append(['1', str(pocet-1)])
		n += 1
	print(n)
	for re in res:
		print(len(re), end = " ") 
		print(" ".join(re))