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
input = sys.stdin.readline
 
def update(operations,c):
    if operations is None:
        return c
    new = []
    for i in operations:
        new.append(c[:i])
        c = c[i:]
    c2 =[]
    for i in range(len(new)-1,-1,-1):
        c2.extend(new[i])
    return c2
    
def prog():
    n = int(input())
    c = list(map(int,input().split()))
    fix_l = 1
    fix_r = n
    tot_moves =[]
    
    for i in range(n//2):
        idxl = c.index(fix_l)
        idxr = c.index(fix_r)
        moves1 = []
        moves2 = []
        
        if idxl > idxr:
            if fix_l > 1:
                moves1.append(fix_l-1)
            moves1.append(n-(fix_l-1) -(n-fix_r))
            if fix_r < n:
                moves1.append(n-fix_r)
            if len(moves1) == 1:
                moves1 = None
        else:
            if fix_l > 1:
                moves1.append(fix_l-1)
            moves1 += [1]*(n-(fix_l-1) -(n-fix_r))
            if fix_r < n:
                moves1.append(n-fix_r)
                
        c = update(moves1,c)
 
        idxl2 = c.index(fix_l)
        idxr2 = c.index(fix_r)
        if fix_r < n:
            moves2.append(n-fix_r)
        moves2.append(idxr2-(n-fix_r) + 1)
        if idxl2-idxr2-1 > 0:
            moves2.append(idxl2-idxr2-1)
        moves2.append(n-idxl2-(fix_l-1))
        if fix_l > 1:
            moves2.append(fix_l-1)
 
        c = update(moves2,c)
        
        if moves1 != None:
            tot_moves.append(moves1)
        tot_moves.append(moves2)
        fix_l += 1
        fix_r -= 1
 
    print(len(tot_moves))
    for move in tot_moves:
        print(len(move),*move)
        
prog()