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
import math 
from decimal import *
def I(): return int(sys.stdin.readline())
def IL(): return list(map(int,sys.stdin.readline().strip().split()))
def ILS(): return list(map(str,sys.stdin.readline().strip().split()))
def check(l):
	res = []
	for ll in l :
		res.extend(ll)
		# print("res  : ",res)
	cnt = 1
	for ll in res :
		if cnt != ll : return ll,False
		cnt+=1
	return -1,True

def get(l,no):
	cnt = 0
	for ll in l :
		if no in ll : return cnt
		cnt+=1
		
def group(l):
	res = []
	i = 0
	while i < len(l):
		val = l[i]
		j = i+1
		while j < len(l):
			nxt = l[j][0]
			prev = val[-1]
			# print(prev,nxt)
			if prev+1 == nxt : 
				val.extend(l[j])
				j+=1
			else : break
		res.append(val)
		# print(val)
		i = j
	return res

def solve():
	n = I()
	l = IL()
	ll = [[x] for x in l]
	ll = group(ll)
	# print(l,ll)
	res = []
	while(True):
		no, cor = check(ll)
		# print(no,cor)
		if cor : break
		sno = get(ll,no)
		# print(sno,"sno")
		ono = get(ll,no-1)
		# print(ono,"ono")
		cur = []
		i = 0
		while i < len(ll):
			if i == sno : 
				val = []
				j = i 
				while j < ono : 
					val.append(ll[j])
					j+=1
				cur.append(val)
				i = j
			else :
				cur.append([ll[i]])
				i+=1
		# print(cur)
		point = []
		for cc in cur :
			cnt =0
			for cxc in cc:
				cnt+=len(cxc)
			point.append(cnt)
		# print("point",point)
		ncur = cur[::-1]
		fcur = []
		for nnc in ncur :
			fcur.extend(nnc)
		# print(fcur,"fcur")
		ll = group(fcur)
		# print(ll,"ll")
		res.append(point)
		# print("===================")
		pass
	print(len(res))
	for rr in res :
		print(len(rr),end=" ")
		for xr in rr :
			print(xr,end=" ")
		print()
solve()