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]. 
from collections import Counter, defaultdict, deque
import bisect
from sys import stdin, stdout
from itertools import repeat
import math

# sys.stdin = open('input')

def mod(x, y, mod):
    re = 1
    now = x
    while y:
        if y&1:
            re *= now
            re %= mod
        y >>= 1
        now = (now*now)%mod
    return re

def inp(force_list=False):
    re = map(int, raw_input().split())
    if len(re) == 1 and not force_list:
        return re[0]
    return re

def inst():
    return raw_input().strip()

def gcd(x, y):
   while(y):
       x, y = y, x % y
   return x

MOD = int(1e9+7)

def query(ty, num):
    if ty==0:
        print 'A %d' % num
    else:
        print 'B %d' % num
    stdout.flush()
    return inp()

def my_main():
    T = 1
    ans = []
    for _ in range(T):
        n = inp()
        da = inp(True)
        op = []
        st = 0
        l = 1
        nl = 0
        mid = n/2+1+n%2
        nm = 0
        for i in range(n-n%2):
            if st==0:
                idx = da.index(mid)
                da = da[n-nm:] + da[idx:n-nm] +  da[nl:idx] + da[0:nl]
                tmpp = filter(lambda x: x!=0, (nl-0, idx-nl, n-nm-idx, n-n+nm))
                len(tmpp) > 1 and ans.append(tmpp)
                mid += 1
                nm += 1
            else:
                idx = da.index(l)
                da = da[n-nl:] + da[idx:n-nl] + da[nm:idx] + da[0:nm]
                tmpp = filter(lambda x: x!=0, (nm-0, idx-nm, n-nl-idx, n-n+nl))
                len(tmpp) > 1 and ans.append(tmpp)
                l += 1
                nl += 1
            # print da, tmpp
            # print l, mid,'kkk',st
            st ^= 1
        print len(ans)
        for iq in ans:
            iq = filter(lambda x: x!=0, iq)
            print len(iq), ' '.join(map(str, iq))










my_main()
