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.*;
 
public class Solution extends PrintWriter {
    
    
    private void solve()  {
    	n = sc.nextInt();
    	c = new int[n];
    	for(int i = 0; i < n; i++) c[i] = sc.nextInt();
    	ArrayDeque<ArrayDeque<Integer>> ans = new ArrayDeque<>();
    	while(true) {
        	ArrayList<ArrayDeque<Integer>> parts = parts();
        	boolean found = false;
        	int fi = 0;
        	int fj = 0;
    		for(int i = 0; i < parts.size() && !found; i++) {
    			ArrayDeque<Integer> p_i = parts.get(i);
    			for(int j = i+1; j < parts.size() && !found; j++) {
    				ArrayDeque<Integer> p_j = parts.get(j);
    				if(p_i.getFirst() == p_j.getLast()+1) {
    					found = true;
    					fi = i;
    					fj = j;
    				}
    			}
    		}
    		if(!found) break;
    		ArrayDeque<Integer> divisions_print = new ArrayDeque<Integer>();
    		ArrayDeque<ArrayDeque<Integer>> invert = new ArrayDeque<>();
    		
    		divide(parts, 0, fi-1, divisions_print, invert);
    		divide(parts, fi, fj-1, divisions_print, invert);
    		divide(parts, fj, fj, divisions_print, invert);
    		divide(parts, fj+1, parts.size()-1, divisions_print, invert);
    		    		
    		ans.add(divisions_print);
    		int idx = 0;
    		for(ArrayDeque<Integer> part : invert) {
    			for(int x : part) {
    				c[idx++] = x;
    			}
    		}
    	}
    	println(ans.size());
    	for(ArrayDeque<Integer> divisions : ans) {
    		print(divisions.size() + " ");
    		for(int x : divisions) print(x + " ");
    		println();
    	}
    }
    
    void divide(ArrayList<ArrayDeque<Integer>> parts, int l, int r, ArrayDeque<Integer> divisions_print, ArrayDeque<ArrayDeque<Integer>> invert){
    	ArrayDeque<Integer> cur =  new ArrayDeque<>();
    	for(int i = l; i <= r; i++) {
			cur.addAll(parts.get(i));
		}
    	if(cur.size() > 0) {
			divisions_print.add(cur.size());
			invert.addFirst(cur);
		}
    }
    
    int[] c;
    int n;
    
    ArrayList<ArrayDeque<Integer>> parts() {
    	ArrayList<ArrayDeque<Integer>> parts = new ArrayList<>();
    	for(int i = 0; i < n; ) {
        	ArrayDeque<Integer> part = new ArrayDeque<>();
        	part.add(c[i]);
    		int j = i+1;
    		while(j < n && c[j-1]+1 == c[j]) {
    			part.add(c[j]);
    			j++;
    		}
    		parts.add(part);
    		i = j;
    	}
    	return parts;
    }
        
//  Solution() throws FileNotFoundException { super(new File("output.txt")); }
//  InputReader sc = new InputReader(new FileInputStream("test_input.txt"));
  Solution() { super(System.out); }
  InputReader sc = new InputReader(System.in);
  static class InputReader {
      InputReader(InputStream in) { this.in = in; } InputStream in;
      
      private byte[] buf = new byte[16384];
      private int    curChar;
      private int    numChars;
      
 
      public int read() {
          if (numChars == -1)
              throw new InputMismatchException();
          if (curChar >= numChars) {
              curChar = 0;
              try {
                  numChars = in.read(buf);
              } catch (IOException e) {
                  throw new InputMismatchException();
              }
              if (numChars <= 0)
                  return -1;
          }
          return buf[curChar++];
      }
 
      public String nextLine() {
          int c = read();
          while (isSpaceChar(c))
              c = read();
          StringBuilder res = new StringBuilder();
          do {
              res.appendCodePoint(c);
              c = read();
          } while (!isEndOfLine(c));
          return res.toString();
      }
 
      public String nextString() {
          int c = read();
          while (isSpaceChar(c))
              c = read();
          StringBuilder res = new StringBuilder();
          do {
              res.appendCodePoint(c);
              c = read();
          } while (!isSpaceChar(c));
          return res.toString();
      }
 
      public long nextLong() {
          int c = read();
          while (isSpaceChar(c))
              c = read();
          int sgn = 1;
          if (c == '-') {
              sgn = -1;
              c = read();
          }
          long res = 0;
          do {
              if (c < '0' || c > '9')
                  throw new InputMismatchException();
              res *= 10;
              res += c - '0';
              c = read();
          } while (!isSpaceChar(c));
          return res * sgn;
      }
 
      public int nextInt() {
          int c = read();
          while (isSpaceChar(c))
              c = read();
          int sgn = 1;
          if (c == '-') {
              sgn = -1;
              c = read();
          }
          int res = 0;
          do {
              if (c < '0' || c > '9')
                  throw new InputMismatchException();
              res *= 10;
              res += c - '0';
              c = read();
          } while (!isSpaceChar(c));
          return res * sgn;
      }
 
      private boolean isSpaceChar(int c) {
          return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
      }
 
      private boolean isEndOfLine(int c) {
          return c == '\n' || c == '\r' || c == -1;
      }
  }
 
    public static void main(String[] $) {
        new Thread(null, new Runnable() {
            public void run() {
                long start = System.nanoTime();
                try {Solution solution = new Solution(); solution.solve(); solution.flush();} 
                catch (Exception e) {e.printStackTrace(); System.exit(1);}
                System.err.println((System.nanoTime()-start)/1E9);
            }
        }, "1", 1 << 27).start();
 
    }
    

}