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:
ans = []

def doRev(arr):
    # print("REVERSE")
    preAns = [len(arr), *([1]*len(arr))]
    ans.append(preAns)
    return list(reversed(arr))

def do(arr, splits):
    splits.append(len(arr))
    splits = list(set(splits))
    splits.sort()
    preAns = [len(splits)]
    l = 0
    for i in range(len(splits)):
        preAns.append(splits[i]-l)
        l = splits[i]
    ans.append(preAns)
    # print(splits, preAns)
    newArr = []
    l = 0
    for i in range(len(splits)):
        newArr.append(arr[l:splits[i]])
        l = splits[i]

    # print(newArr, "How we split it")
    newArr.reverse()
    new = []
    for i in range(len(newArr)):
        for g in range(len(newArr[i])):
            new.append(newArr[i][g])
    # print(new, "what we got")
    # print()
    return new

def doinotherway(arr):
    n = len(arr)
    arr = arr
    it = 0
    while it < n+10:
        it += 1
        f = arr[0]
        if f == 1:
            if arr == list(reversed(list(sorted(arr)))):
                arr = doRev(arr)
            else:
                more = 0
                for i in range(1, n):
                    if arr[i] == arr[i - 1] + 1:
                        more += 1
                    else:
                        break
                arr = do(arr, [more + 1])
        else:
            more = 0
            for i in range(1, n):
                if arr[i] != arr[i - 1] + 1:
                    break
                else:
                    more += 1
            ind = 0
            for i in range(n):
                if arr[i] == f - 1:
                    ind = i
                    break
            more2 = 0
            for i in range(ind - 1, -1, -1):
                if arr[i] != arr[i + 1] - 1:
                    break
                else:
                    more2 += 1

            arr = do(arr, [ind - more2, ind + 1])

        if arr == list(reversed(list(sorted(arr)))):
            arr = doRev(arr)
        if arr == list(sorted(arr)):
            break
        # print(arr,'NEW ARRAY---------------------')
    # print(arr, 'FIN')
    # print(arr)

def donormal(arr):
    n = len(arr)
    arr = arr
    it = 0
    while it < n+10:
        it += 1
        f = arr[0]
        if f == n:
            if arr == list(reversed(list(sorted(arr)))):
                arr = doRev(arr)
            else:
                more = 0
                for i in range(1, n):
                    if arr[i] == arr[i - 1] - 1:
                        more += 1
                    else:
                        break
                arr = do(arr, [more + 1])
        else:
            more = 0
            for i in range(1, n):
                if arr[i] != arr[i - 1] - 1:
                    break
                else:
                    more += 1
            ind = 0
            for i in range(n):
                if arr[i] == f + 1:
                    ind = i
                    break
            more2 = 0
            for i in range(ind - 1, -1, -1):
                if arr[i] != arr[i + 1] + 1:
                    break
                else:
                    more2 += 1

            arr = do(arr, [ind - more2, ind + 1])

        if arr == list(reversed(list(sorted(arr)))):
            arr = doRev(arr)
        if arr == list(sorted(arr)):
            break
        # print(arr,'NEW ARRAY---------------------')
    # print(arr, 'FIN')
    # print(arr)

def solve():
    n = int(input())
    arr = list(map(int,input().split()))
    c = arr[:]
    if n == 1 or arr == list(sorted(arr)):
        print(0)
        return 0
    donormal(arr)
    global ans
    if len(ans) <=n:
        print(len(ans))
        for i in range(len(ans)):
            print(*ans[i])
    else:
        ans = []
        doinotherway(c)
        print(len(ans))
        for i in range(len(ans)):
            print(*ans[i])

solve()