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
import bisect
from sys import stdin, stdout
from math import gcd, floor, sqrt, log2, ceil
from collections import defaultdict as dd
from bisect import bisect_left as bl, bisect_right as br
from bisect import insort
from collections import Counter
from collections import deque
from heapq import heappush,heappop,heapify
from itertools import permutations,combinations
mod = int(1e9)+7
 
 
ip = lambda : int(stdin.readline())
inp = lambda: map(int,stdin.readline().split())
ips = lambda: stdin.readline().rstrip()
out = lambda x : stdout.write(str(x)+"\n")
 
t = 1
for _ in range(t):
    n = ip()
    arr = list(inp())
    perm = [i for i in range(1,n+1)]
    ct = 1
    pt = 0
    sub = [arr[0]]
    ans = []
    while len(sub) != n:
        k = len(sub)
        part = [0]*n
        element = arr[k]
        for i in range(n-k+1):
            if sorted(list(perm[i:i+k])) == sorted(list(sub)):
                start = i
                end = i+k
                ch = list(perm[i:i+k])
                for i in range(start,end):
                    part[i] = 1
                if ch == sub:
                    pass
                else:
                    temp = list(ch)
                    ele = temp.pop(0)
                    temp.append(ele)
                    if temp == sub:
                        part[start] = 2
                    else:
                        part[end-1] = 2
                break
        flag = False
        for i in range(start-1,-1,-1):
            if perm[i] == element:
                flag = True
                break
        if flag:
            for i in range(start-1,-1,-1):
                if perm[i] == element:
                    part[i] = 3
                    break
                else:
                    part[i] = 3
        else:
            for i in range(end,n):
                if perm[i] == element:
                    part[i] = 3
                    break
                else:
                    part[i] = 3
        sub.append(element)
        prev = None
        ct = 1
        stack = []
        gap = []
        for i in part:
            if len(stack) == 0:
                stack.append(i)
            else:
                if stack[-1] == i:
                    stack.append(i)
                else:
                    gap.append(len(stack))
                    stack = [i]
        gap.append(len(stack))
        g2 = list(gap)
        g2 = g2[::-1]
        g2 = [len(g2)]+g2
        final = " ".join([str(i) for i in g2])
        part2 = []
        st = 0
        for i in gap:
            temp = perm[st:st+i]
            part2.append(temp)
            st += i
        ans.append(final)
        part2 = part2[::-1]
        perm = []
        for i in part2:
            perm += i
    if perm == arr:
        pass
    else:
        ele = perm.pop(0)
        perm.append(ele)
        if arr == perm:
            ans.append("{} {} {}".format(2,n-1,1))
        else:
            ans.append("{} {} {}".format(2,1,n-1))
            ele = perm.pop()
            perm = [ele] + perm
    ans = ans[::-1]
    print(len(ans))
    for i in ans:
        print(i)
        
                
            
        
        
        
