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.io.*;
import java.util.*;
 
 
//Sub_To_Errichto
public class Main {
    static final long MOD = 1000000007L;
    //static final long MOD = 998244353L;
    static final int INF = 50000000;
    static final int NINF = -500000000;
    
    static ArrayList<ArrayList<Integer>> ops;
    public static void main(String[] args) {
        FastScanner sc = new FastScanner();
        PrintWriter pw = new PrintWriter(System.out);
        
        int N = sc.ni();
        ArrayList<Block> nums = new ArrayList<Block>();
        for (int i = 0; i < N; i++) {
        	nums.add(new Block(sc.ni(),1));
        }
        nums = merge(nums);
       
        ops = new ArrayList<ArrayList<Integer>>();
        while (nums.size() > 1) {
        	int S = nums.size();
        	int[] swap = new int[2];
        	Arrays.fill(swap,-1);
        	loop:
        	for (int i = 0; i < S; i++) {
        		for (int j = i+1; j < S; j++) {
        			if (nums.get(j).end == nums.get(i).end-nums.get(i).size) {
        				swap = new int[] {i,j};
        				break loop;
        			}
        		}
        	}
        	if (swap[0] == 0 && swap[1] == S-1) {
        		nums = op(nums,new int[] {0,swap[1]});
        	} else if (swap[0] == 0) {
        		nums = op(nums,new int[] {0,swap[1],swap[1]+1});
        	} else if (swap[1]==S-1) {
        		nums = op(nums,new int[] {0,swap[0],swap[1]});
        	} else {
        		nums = op(nums,new int[] {0,swap[0],swap[1],swap[1]+1});
        	}
        }
        pw.println(ops.size());
        for (ArrayList<Integer> op: ops) {
        	pw.print(op.size() + " ");
        	for (int a: op)
        		pw.print(a + " ");
        	pw.println();
        }
        
        pw.close();
    }
    
    public static ArrayList<Block> op(ArrayList<Block> nums, int[] cut) {
    	ArrayList<Block> ans = new ArrayList<Block>();
    	ArrayList<Integer> op = new ArrayList<Integer>();
    	int last = nums.size()-1;
    	for (int i = cut.length-1; i >= 0; i--) {
    		int sz = 0;
    		for (int j = cut[i]; j <= last; j++) {
    			ans.add(nums.get(j));
    			sz += nums.get(j).size;
    		}
    		op.add(sz);
    		last = cut[i]-1;
    	}
    	Collections.reverse(op);
    	ops.add(op);
    	ans = merge(ans);
    	return ans;
    }
    
    public static ArrayList<Block> merge(ArrayList<Block> nums) {
    	ArrayList<Block> ans = new ArrayList<Block>();
    	Block cur = nums.get(0);
    	for (int i = 1; i < nums.size(); i++) {
    		if (nums.get(i).end-nums.get(i).size == cur.end) {
    			cur = new Block(nums.get(i).end,nums.get(i).size+cur.size);
    		} else {
    			ans.add(cur);
    			cur = nums.get(i);
    		}
    	}
    	ans.add(cur);
    	return ans;
    }
    
    static class Block {
    	int size;
    	int end;
    	public Block(int e, int s) {
    		size = s;
    		end = e;
    	}
    	
    	public String toString() {
    		return "["+end+","+size+"]";
    	}
    }
    
    static class FastScanner {
        BufferedReader br;
        StringTokenizer st;
 
        public FastScanner() {
            br = new BufferedReader(new InputStreamReader(System.in));
        }
 
        String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }
 
        int ni() {
            return Integer.parseInt(next());
        }
 
        long nl() {
            return Long.parseLong(next());
        }
 
        double nd() {
            return Double.parseDouble(next());
        }
 
        String nextLine() {
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
    }
}