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]. 
n = int(input())
c = list(map(int, input().split()))
q = 0
k = []
d = []
x = 0
arr=[]
if c[0] != 1:
    q += 1
    k.append(2)
    for i in range(n):
        if c[i] == 1:
            x = i
            break
    arr=[x, n-x]
    d.append([x, n - x])
    s = [0]
    for j in range(len(arr) - 1):
        s.append(s[-1] + arr[j])
    s.append(n)
    c1 = []
    for j in range(len(s) - 1, 0, -1):
        c1 += c[s[j - 1]:s[j]]
    c = c1
f = True
t = 1
for i in range(n - 2):
    if f:
        arr = []
        for _ in range(t):
            arr.append(1)
        for j in range(t, n):
            if c[j] == t + 1:
                x = j
                break
        arr.append(x - t + 1)
        if (n > (x + 1)):
            arr.append(n - (x + 1))
            k.append(t + 2)
        else:
            k.append(t + 1)
        d.append(arr)
        s = [0]
        for j in range(len(arr) - 1):
            s.append(s[-1] + arr[j])
        s.append(n)
        c1 = []
        for j in range(len(s) - 1, 0, -1):
            c1 += c[s[j - 1]:s[j]]
        c = c1
        t += 1
        f=False
    else:
        arr = []
        for j in range(n):
            if c[j] == t + 1:
                x = j
                break
        if (x > 0):
            arr.append(x)
            arr.append(n - x - t)
            k.append(t + 2)
        else:
            arr.append(n - t)
            k.append(t + 1)
        for _ in range(t):
            arr.append(1)
        d.append(arr)
        t += 1
        s = [0]
        for j in range(len(arr) - 1):
            s.append(s[-1] + arr[j])
        s.append(n)
        c1 = []
        for j in range(len(s) - 1, 0, -1):
            c1 += c[s[j - 1]:s[j]]
        c = c1
        f=True
if c[0]!=1:
    arr=[]
    for _ in range(n):
        arr.append(1)
    d.append(arr)
    k.append(n)
print(len(k))
for i in range(len(k)):
    print(k[i], end=' ')
    print(*d[i])
