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]. 
import java.io.BufferedOutputStream;
import java.io.BufferedReader;
import java.io.FileNotFoundException;
import java.io.FileReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayDeque;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.HashMap;
import java.util.HashSet;
import java.util.InputMismatchException;
import java.util.Map;
import java.util.Queue;
import java.util.Random;
import java.util.Stack;
import java.util.StringTokenizer;
import java.util.TreeMap;
 
public class Main {
	
public static void main(String[] args){
 
   new Thread(null, null, "Anshum Gupta", 99999999) {
        public void run() {
            try {
                solve();
            } catch(Exception e) {
                e.printStackTrace();
                System.exit(1);
            }
        }
    }.start();
}
 
static final long mxx = (long)(1e18 + 5);
static final int mxN = (int)(1e6);
static final int mxV = (int)(1e6), log = 18;
static final long MOD = 998244353; 
static final int INF = (int)1e9;
static boolean[]vis;
static ArrayList<ArrayList<Integer>> adj;
static int n, m, k, q, count = 0;
static char[]str;
static StringBuilder ans = new StringBuilder();

public static void solve() throws Exception {
	   // solve the problem here
		s = new MyScanner();
   		out = new PrintWriter(new BufferedOutputStream(System.out), true);
//	   		out = new PrintWriter("output.txt");
        int tc = 1;//s.nextInt();
        for(int i=1; i<=tc; i++) {
//        	out.print("Case #" + i + ": ");
        	testcase();
        }
           
        out.flush();
        out.close();
}


static void testcase() {
	n = s.nextInt();
	int[] c = s.nextIntArray(n);
	boolean inc = true;
	for(int i = 1; i <= n; i++) {
		// move i to correct position
		int index = findIndex(c, i);
		if(i == 1) {
			if(index == 0)continue;
			ArrayList<Integer> cur = new ArrayList<Integer>();
			if(n % 2 == 0) {		
				cur.add(index + 1);
				if(n - index - 1 > 0)
				cur.add(n - index - 1);
				print(cur);
				makeChange(c, cur, inc);
				inc = false;
			}else {
				if(index > 0)
				cur.add(index);
				cur.add(n - index);
				print(cur);
				makeChange(c, cur, inc);
			}
		}else {
			if(inc) {
				ArrayList<Integer> cur = new ArrayList<Integer>();
				for(int j = 1; j < i; j++) {
					cur.add(1);
				}
				cur.add(index - i + 2);
				if(n - index - 1 > 0)cur.add(n - index - 1);
				print(cur);
				makeChange(c, cur, inc);
				inc = false;
			}else {
				ArrayList<Integer> cur = new ArrayList<Integer>();
				for(int j = 1; j < i; j++) {
					cur.add(1);
				}
				cur.add(n - i - index + 1);
				if(index > 0)cur.add(index);
				Collections.reverse(cur);
				print(cur);
				makeChange(c, cur, inc);
				inc = true;
			}
		}
//		out.println("i = " + i + " c => " + Arrays.toString(c));
	}
	if(!inc) {
		ArrayList<Integer> cur = new ArrayList<Integer>();
		for(int j = 0; j < n; j++) cur.add(1);
		print(cur);
		makeChange(c, cur, false);
//		out.println(Arrays.toString(c));
	}
	out.println(count);
	out.println(ans);
}



private static void makeChange(int[] c, ArrayList<Integer> cur, boolean inc) {
	int[] copy = new int[n];
	if(inc) {
		int last = n, start = -1;
		for(int len : cur) {
			for(int j = last - len; j < last; j++) {
				copy[j] = c[++start];
			}
			last -= len;
		}
	}else {
		Collections.reverse(cur);
		int last = -1, start = n;	
		for(int len : cur) {
			for(int j = last + len; j > last; j--) {
				copy[j] = c[--start];
			}
			last += len;
		}
	}
	for(int j = 0; j < n; j++) c[j] = copy[j];
	
}


private static void print(ArrayList<Integer> cur) {
	if(cur.size() <= 1)return;
	count++;
	ans.append(cur.size() + " ");
	for(int x : cur)ans.append(x + " ");
	ans.append("\n");
	
}


private static int findIndex(int[] c, int i) {
	for(int j = 0; j < n; j++) {
		if(c[j] == i)return j;
	}
	throw null;
}

public static PrintWriter out;
public static MyScanner s;
static void shuffleArray(int[] a) {
	Random random = new Random();
	for (int i = a.length-1; i > 0; i--) {
		int index = random.nextInt(i + 1);
		int tmp = a[index];
		a[index] = a[i];
		a[i] = tmp;
	}
}
static void shuffleSort(int[] a) {
	shuffleArray(a);
	Arrays.parallelSort(a);
}
static class MyScanner {
 
    BufferedReader br;
    StringTokenizer st;
 
    public MyScanner() {
        br = new BufferedReader(new InputStreamReader(System.in));
    }
    public MyScanner(String fileName) {
    	try {
			br = new BufferedReader(new FileReader(fileName));
		} catch (FileNotFoundException e) {
			e.printStackTrace();
		}
    }
    String next() {
        while (st == null || !st.hasMoreElements()) {
            try {
                st = new StringTokenizer(br.readLine());
            } catch (IOException e) {
                e.printStackTrace();
            }
        }
        return st.nextToken();
    }
 
    int nextInt() { return Integer.parseInt(next()); }
    long nextLong() { return Long.parseLong(next()); }
    double nextDouble() { return Double.parseDouble(next()); }
    int[] nextIntArray(int n){
    	int[]a = new int[n];
    	for(int i=0; i<n; i++) {
    		a[i] = this.nextInt();
    	}
    	return a;
    }
    long[] nextlongArray(int n) {
    	long[]a = new long[n];
    	for(int i=0; i<n; i++) {
    		a[i] = this.nextLong();
    	}
    	return a;
    }
    Integer[] nextIntegerArray(int n){
    	Integer[]a = new Integer[n];
    	for(int i=0; i<n; i++) {
    		a[i] = this.nextInt();
    	}
    	return a;
    }
    Long[] nextLongArray(int n) {
    	Long[]a = new Long[n];
    	for(int i=0; i<n; i++) {
    		a[i] = this.nextLong();
    	}
    	return a;
    }
    char[][] next2DCharArray(int n, int m){
    	char[][]arr = new char[n][m];
    	for(int i=0; i<n; i++) {
    		arr[i] = this.next().toCharArray();
    	}
    	return arr;
    }
    ArrayList<ArrayList<Integer>> readUndirectedUnweightedGraph(int n, int m) {
    	ArrayList<ArrayList<Integer>>adj = new ArrayList<ArrayList<Integer>>();
    	for(int i=0; i<=n; i++)adj.add(new ArrayList<Integer>());
    	for(int i=0; i<m; i++) {
    		int u = s.nextInt();
    		int v = s.nextInt();
    		adj.get(u).add(v);
    		adj.get(v).add(u);
    	}
    	return adj;
    }
    ArrayList<ArrayList<Integer>> readDirectedUnweightedGraph(int n, int m) {
    	ArrayList<ArrayList<Integer>>adj = new ArrayList<ArrayList<Integer>>();
    	for(int i=0; i<=n; i++)adj.add(new ArrayList<Integer>());
    	for(int i=0; i<m; i++) {
    		int u = s.nextInt();
    		int v = s.nextInt();
    		adj.get(u).add(v);
    	}
    	return adj;
    }
    String nextLine(){
        String str = "";
        try {
            str = br.readLine();
        } catch (IOException e) {
            e.printStackTrace();
        }
        return str;
    }
}
 
 
}