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 sys import stdin
input=stdin.readline

def A():
    t=int(input())
    for _ in range(t):
        n=int(input())
        a=list(map(int,input().split()))
        if sum(a) == 0:
            print("NO")
        elif sum(a) > 0:
            a.sort(reverse=True)
            print("YES")
            print(*a)
        else:
            a.sort()
            print("YES")
            print(*a)

def B():
    from collections import deque
    t=int(input())
    for _ in range(t):
        n,k=map(int,input().split())
        s=[bool(ord(x) - ord('L')) for x in input().rstrip()]

        lastGame = False
        score = 0
        for i in range(n):
            if s[i]:
                if lastGame:
                    score += 2
                else:
                    score += 1
                lastGame=True
            else:
                lastGame=False

        startW=0
        endW=n-1
        while startW < n and not s[startW]:
            startW += 1
        while endW >= 0 and not s[endW]:
            endW -= 1
        streak=0
        wlws=deque()
        for i in range(startW+1, endW+1):
            if s[i] and streak > 0:
                wlws.append(streak)
                streak=0
            elif not s[i]:
                streak += 1
        
        wlws=sorted(list(wlws))

        j = 0
        while j < len(wlws) and k >= wlws[j]:
            score += (2 * wlws[j] + 1)
            k -= wlws[j]
            j += 1
            
        if k > 0:
            remainingLs = startW + (n - 1 - endW)
            if startW == n:
                remainingLs = n
                score -= 1
            if j < len(wlws):
                remainingLs = wlws[j]
            score += min(remainingLs, k) * 2
        print(score)
        

def C():
    r,n=map(int,input().split())
    xs=[0]*n
    ys=[0]*n
    ts=[0]*n
    tots=[0]*n
    for i in range(n):
        ts[i],xs[i],ys[i]=map(int,input().split())
    for i in range(n - 1, -1, -1):
        j = i + 1
        while j < n and ts[j]-ts[i] <= 4*r:
            if abs(xs[j]-xs[i])+abs(ys[j]-ys[i]) <= ts[j]-ts[i]:
                tots[i] = max(tots[j], tots[i])
            j += 1
        if xs[i]-1 + ys[i]-1 <= ts[i]:
            tots[i] += 1
        else:
            tots[i]=0
    print(max(tots))
        

def D():        
    n=int(input())
    c=list(map(int,input().split()))

    def transform(parts,c):
        h=[]
        i=0
        for part in parts:
            h.append(c[i:i+part])
            i += part
        c.clear()
        for j in range(len(h) - 1, -1, -1):
            c += h[j]
        return c

    if c == [1] or c == sorted(c):
        print(0)
        return
    transforms=[[] for i in range(52)]
    targetInd=c.index(1)
    if targetInd != 0:
        transforms[0]=[2,targetInd,n-targetInd]
        c=transform([targetInd,n-targetInd],c)
        transIndex=1
    else:
        transIndex=0
    currTrans=[]

    numSorted=1
    targetNum=2
    inverted=False
    
    while c != sorted(c):
        if not inverted:
            k=c.index(targetNum) + 1 - numSorted
            leftOver=n - numSorted - k
            if leftOver != 0:
                currTrans = ([1]*numSorted) + [k] + [leftOver]
            else:
                currTrans = ([1]*numSorted) + [k]
        else:
            k=n - numSorted - c.index(targetNum)
            leftOver=n - numSorted - k
            if leftOver != 0:
                currTrans = [leftOver] + [k] + ([1]*numSorted)
            else:
                currTrans = [k] + ([1]*numSorted)
        c=transform(currTrans,c)
        numSorted += 1
        targetNum += 1
        inverted = not inverted
        transforms[transIndex] = [len(currTrans)] + currTrans
        transIndex += 1

    print(transIndex)
    for i in range(transIndex):
        print(*transforms[i])        
    
D()
