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 class Point{
		int x,y;
		Point(int nx, int ny){x = nx; y = ny;}
	}
	public static void main(String[] args) throws IOException{
		BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
		int n = pint(in);
		int[] a = new int[n];
		StringTokenizer st = new StringTokenizer(in.readLine());
		for(int i = 0; i < n; i++) {
			a[i] = pint(st);
		}
		int t = 0;
		int k = 0;
		for(int i = 0; i < n; i++) {
			if(a[i] == 1) {
				k = i + 1;
				break;
			}
		}
		StringBuilder sb = new StringBuilder();
		ArrayList<Integer> c = new ArrayList<Integer>();

		if(k != n) {
			sb.append(2);
			sb.append(' ');
			sb.append(k);
			sb.append(' ');
			sb.append(n - k);
			sb.append('\n');
			t = 1;
			c.add(n - k);
			c.add(k);
			a = swap(a,c);
		}
		int sf = 1;
		boolean h = false;
		for(int i = 2; i <= n; i += 2) {
			c.clear();
			k = 0;
			for(int j = 0; j < n; j++) {
				if(a[j] == i) {k = j; break;}
			}
			c.add(sf);
			if(h) {
				c.add(1);
				if(n - sf - k - 1!= 0) {c.add(n - sf - k - 1);}
			}else {
				if(n - sf - k != 0) {c.add(n - sf - k);}
			}
			if(k != 0) {c.add(k);}
			t++;
			sb.append(c.size());
			for(int j = c.size() - 1; j >= 0; j--) {
				sb.append(' ');
				sb.append(c.get(j));
			}
			sb.append('\n');
			a = swap(a,c);
			sf++;
			if(h) {sf++;}
			if(i + 1 >= n) {break;}
			
			c.clear();
			for(int j = sf - 1; j < n; j++) {
				if(a[j] == i + 1) {k = j; break;}
			}
			k -= (sf - 1);
			if(n - sf - k != 0) {c.add(n - sf - k);}
			if(k != 0) {c.add(k);}
			c.add(sf);
			a = swap(a,c);
			t++;
			sb.append(c.size());
			for(int j = c.size() - 1; j >= 0; j--) {
				sb.append(' ');
				sb.append(c.get(j));
			}
			sb.append('\n');
			h = true;
		}
		System.out.println(t);
		System.out.print(sb);
	}
	static int[] swap(int[] a, ArrayList<Integer> c) {
		int[] b = new int[a.length];
		int p = 0;
		for(int i : c) {
			System.arraycopy(a, a.length - p - i, b, p, i);
			p += i;
		}
		return b;
	}
    static int pint(BufferedReader in) throws IOException {return Integer.parseInt(in.readLine());}
    static int pint(StringTokenizer st) {return Integer.parseInt(st.nextToken());}
}