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]. 
import sys
#from collections import deque
#from functools import *
#from fractions import Fraction as f
#from copy import *
#from bisect import *	
#from heapq import *
#from math import gcd,ceil,sqrt
#from itertools import permutations as prm,product
 
def eprint(*args):
    print(*args, file=sys.stderr)
zz=1
 
#sys.setrecursionlimit(10**6)
if zz:
	input=sys.stdin.readline
else:	
	sys.stdin=open('input.txt', 'r')
	sys.stdout=open('all.txt','w')
di=[[-1,0],[1,0],[0,1],[0,-1]]

def string(s):
	return "".join(s)
def fori(n):
	return [fi() for i in range(n)]	
def inc(d,c,x=1):
	d[c]=d[c]+x if c in d else x
def bo(i):
	return ord(i)-ord('A')	
def li():
	return [int(xx) for xx in input().split()]
def fli():
	return [float(x) for x in input().split()]	
def comp(a,b):
	if(a>b):
		return 2
	return 2 if a==b else 0		
def gi():	
	return [xx for xx in input().split()]
def cil(n,m):
	return n//m+int(n%m>0)	
def fi():
	return int(input())
def pro(a): 
	return reduce(lambda a,b:a*b,a)		
def swap(a,i,j): 
	a[i],a[j]=a[j],a[i]	
def si():
	return list(input().rstrip())	
def mi():
	return 	map(int,input().split())			
def gh():
	sys.stdout.flush()
def isvalid(i,j):
	return 0<=i<n and 0<=j<m and a[i][j]!="."
def bo(i):
	return ord(i)-ord('a')	
def graph(n,m):
	for i in range(m):
		x,y=mi()
		a[x].append(y)
		a[y].append(x)


t=1
while t>0:
	t-=1
	n=fi()
	a=li()
	ans=[]
	for _ in range(n):
		vis={}
		for i in range(n):
			if a[i]+1 in vis:
				j=i-1
				while j>=0 and a[j]==a[j+1]-1:
					j-=1
				#print(i,j,a,a[j])	
				p=[vis[a[i]+1],j-vis[a[i]+1]+1,i-j,n-i-1]
				if p[0]==0:
					p=p[1:]
				if p[-1]==0:
					p=p[:-1]
				ans.append([len(p)]+p)	
				break
			vis[a[i]]=i
		d=[]
		c=0
		if a==sorted(a):
			break
		for j in p:
			d.append(a[c:c+j])
			c=c+j
		a=[]
		for i in range(len(d)-1,-1,-1):
			a+=d[i]	
	print(len(ans))		
	for i in ans:
		print(*i)					
