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:
import sys
from collections import deque
n = int(input())
arr = list(map(int , sys.stdin.readline().strip().split()))
ans = []
while(True):
    idx_kel_akhir = n - 1
    while idx_kel_akhir > 0 and arr[idx_kel_akhir - 1] == arr[idx_kel_akhir] - 1:
        idx_kel_akhir -= 1
    if idx_kel_akhir == 0:
        break
    if arr[-1] != n:
        idx_pasang = idx_kel_akhir - 1
        while arr[idx_pasang] != arr[-1] + 1:
            idx_pasang -= 1
        if idx_pasang == 0:
            ans.append([idx_kel_akhir , n - idx_kel_akhir])
        else:
            ans.append([idx_pasang , idx_kel_akhir - idx_pasang , n - idx_kel_akhir])
    else:
        idx_kel_akhir_dua = idx_kel_akhir - 1
        while arr[idx_kel_akhir_dua - 1] == arr[idx_kel_akhir_dua] - 1:
            idx_kel_akhir_dua -= 1
        idx_pasang = idx_kel_akhir_dua - 1
        while arr[idx_pasang] != arr[idx_kel_akhir - 1] + 1:
            idx_pasang -= 1
        #print("idx kel akhir" , idx_kel_akhir)
        #print("idx_kel_akhir_dua" , idx_kel_akhir_dua)
        #print("idx_pasang" , idx_pasang)
        if idx_pasang == 0:
            ans.append([idx_kel_akhir_dua , idx_kel_akhir - idx_kel_akhir_dua , n - idx_kel_akhir])
        else:
            ans.append([idx_pasang , idx_kel_akhir_dua - idx_pasang , idx_kel_akhir - idx_kel_akhir_dua , n - idx_kel_akhir])
    sem = deque([])
    batas_idx = int(0)
    for tambah in ans[-1]:
        bef = batas_idx
        batas_idx += tambah
        for i in range(batas_idx - 1 , bef - 1 , -1):
            sem.appendleft(arr[i])
    for i in range(n):
        arr[i] = sem[i]
    #print("hasil sekarang")
    #for i in ans[-1]:
    #    sys.stdout.write(str(i) + ' ')
    #print("\nmenjadi")
    #for i in arr:
    #    sys.stdout.write(str(i) + ' ')
    #print()
    
print(len(ans))
for L in ans:
    sys.stdout.write(str(len(L)))
    for i in L:
        sys.stdout.write(' ' + str(i))
    print()