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 JAVA):

Solution:
import java.util.*;
import java.io.*;
public class Main {
    static Scanner sc = new Scanner(System.in);
    static PrintWriter out = new PrintWriter(System.out);
    static BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
    long mod = 998244353;

    public static void main(String[] args){
        Main main = new Main();
        int t = 1;
        while(t-->0){
            main.solve();
        }
        out.flush();
    }
    void solve(){
        int n = sc.nextInt();
        int[] a = new int[n];
        for(int i=0; i<n; i++) a[i] = sc.nextInt();
        if(n==1){
            out.println(0); return;
        }
        List<Deque<Integer>> ans = new ArrayList<>();
        for(int step=0; step<n/2; step++){
            // trying to settle step+1 and n+1-step
            int idxSmall = findIdx(a,step+1), idxLarge = findIdx(a, n-step);
            List<Integer> list = new ArrayList<>();
            Deque<Integer> temp;
            if(idxSmall<idxLarge){
                list.add(step);list.add(idxSmall-step+1);
                list.add(idxLarge-idxSmall-1);
                list.add(n-step-idxLarge); list.add(step);
                changeArr(list, a);
                temp = getOrder(list);
                if(temp.peek()>1) ans.add(temp);

                list.clear();
                list.add(step);list.add(1);list.add(n-2-2*step);list.add(1);list.add(step);
                changeArr(list, a);
                temp = getOrder(list);
                if(temp.peek()>1) ans.add(temp);
            } else{
                list.add(step); list.add(n-2*step); list.add(step);
                changeArr(list, a);
                temp = getOrder(list);
                if(temp.peek()>1) ans.add(temp);

                list.clear();
                list.add(step); list.add(idxLarge-step+1);
                list.add(idxSmall-idxLarge-1);
                list.add(n-step-idxSmall);list.add(step);
                changeArr(list, a);
                temp = getOrder(list);
                if(temp.peek()>1) ans.add(temp);
            }
        }
        out.println(ans.size());
        for(Deque<Integer> de:ans){
            while(de.size()>0){
                out.print(de.pollFirst()+" ");
            }
            out.println();
        }

    }
    int findIdx(int[] a, int target){
        for(int i=0; i<a.length; i++){
            if(a[i]==target) return i;
        }
        return 0;
    }
    void changeArr(List<Integer> list, int[] a){
        int n = a.length, left = 0, end = n-1;
        int[] b = new int[n];
        for(int num:list){
            if(num<=0) continue;
            int start = end+1-num;
            for(int i=0; i<num; i++) b[start+i] = a[left++];
            end = start-1;
        }
        System.arraycopy(b, 0, a, 0, n);
    }
    Deque<Integer> getOrder(List<Integer> list){
        Deque<Integer> ans = new LinkedList<>();
        for(int i:list){
            if(i>0) ans.addLast(i);
        }
        ans.addFirst(ans.size());
        return ans;
    }

}