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:
n = int(input())
l = list(map(int,input().split()))
def so(x):
    for i in range(n-1):
        if l[i+1] < l[i]:
            return False
    return True
op = 0
odp = []
while not so(l):
    if op%2 == 0:
        chuj = 0
        for i in range(n):
            if l[i] == op+1:
                chuj = i
        nowa = [chuj,n-chuj-op]+[1]*op
        if chuj == 0:
            nowa.pop(0)
        if n-chuj-op == 0:
            nowa.pop(1)
        odp.append(nowa)
        dupa = [0]
        for i in nowa:
            dupa.append(dupa[-1]+i)
        podz = []
        k = len(dupa)
        for i in range(k-1):
            podz.append(l[dupa[i]:dupa[i+1]])
        dup = []
        podz.reverse()
        for i in podz:
            dup += i
        l = dup
        #print(l)
    else:
        chuj = 0
        for i in range(n):
            if l[i] == op+1:
                chuj = i
        if chuj < n-1:
            nowa = [1]*op + [chuj-op+1,n-chuj-1]
        else:
            nowa = [1]*op + [chuj-op+1]
        odp.append(nowa)
        dupa = [0]
        for i in nowa:
            dupa.append(dupa[-1]+i)
        podz = []
        k = len(dupa)
        for i in range(k-1):
            podz.append(l[dupa[i]:dupa[i+1]])
        dup = []
        podz.reverse()
        for i in podz:
            dup += i
        l = dup
        #print(l)
    op += 1
odpp = []
for i in odp:
    if len(i) > 1:
        odpp.append(i)
print(len(odpp))
for i in odpp:
    print(len(i), *i)