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:
def shuff(splitarray,c):#break before splitarray's indexes
    temp=[]
    temp2=[]
    for i in range(len(c)):
        if i in splitarray:
            temp.append(temp2)
            temp2=[]
        temp2.append(c[i])
    temp.append(temp2)
    c2=[]
    for te in reversed(temp):
        for i in te:
            c2.append(i)
    return c2

def put2IndexesTogether(idx1L,idx1R,idx2L,idx2R): #put idx2L right after idx1R. return splitarray
    splitarray=[idx2L,idx2R+1,idx1R+1]
    splitarray2=[]
    for i in splitarray:
        if 1<=i<n:
            splitarray2.append(i)
    return splitarray2

def convertSplitArray(splitarray): #return sizes=[size1,size2,...]
    splitarray.append(n)
    latest=0
    sizes=[]
    for x in splitarray:
        sizes.append(x-latest)
        latest=x
    sizes2=[]
    for sz in sizes:
        if sz>0:
            sizes2.append(sz)
    return sizes2
    

n=int(input())
c=[int(x) for x in input().split()]
#n=20
#c=[int(x) for x in '1 6 15 9 18 17 7 8 3 19 2 13 11 12 14 4 5 20 16 10'.split()]

splitarrays=[]
sortedIs=set() #i+1 is already after i
while True:
    idxof=dict()
    for i in range(n):
        idxof[c[i]]=i
    ok=True
    for i in range(1,n):
        if i not in sortedIs and idxof[i+1]<idxof[i]:
            sortedIs.add(i)
            
            idx1R=idxof[i]
            idx1L=idx1R
            while idx1L-1>=0 and c[idx1L-1]+1==c[idx1L]:
                idx1L-=1
            
            idx2L=idxof[i+1]
            idx2R=idx2L
            while idx2R+1<n and c[idx2R+1]==c[idx2R]+1:
                idx2R+=1
            
            splitarray=put2IndexesTogether(idx1L,idx1R,idx2L,idx2R)
            splitarrays.append(splitarray)
#            print(i,sortedIs,c)####
            c=shuff(splitarray,c)
            ok=False
            break
    if ok:break
ans=[]
for x in splitarrays:
    ans.append(convertSplitArray(x))

print(len(ans))
for x in ans:
    s=str(len(x))
    s=s+' '+' '.join([str(y) for y in x])
    print(s)