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.*;
import java.util.*;
import java.math.*;

public class D {
	public static void main(String[] args) throws IOException {

		/**/
		Scanner sc = new Scanner(new BufferedReader(new InputStreamReader(System.in)));
		/*/
		Scanner sc = new Scanner(new BufferedReader(new InputStreamReader(new FileInputStream("src/d.in"))));
		/**/
		
		int n = sc.nextInt();
		int[] c = new int[n];
		for (int i = 0; i < n; ++i)
			c[i] = sc.nextInt()-1;
		
		boolean end = true;
		ArrayList<String> lines = new ArrayList<>();
		for (int i = 0; i < n; ++i) {
			boolean isSorted = true;
			for (int j = 0; j < n; ++j) {
				if (c[j] != j)
					isSorted = false;
			}
			if (isSorted)
				break;
			int ct = 0;
			int pc = 0;
			String ans = "";
			if (end) {
				end = !end;
				for (int j = n-1; j >= 0; --j) {
					if (j+c[j]==n-1)
						++ct;
					else
						break;
				}
				if (ct==n) {
					for (int j = 0; j < n; ++j) {
						ans += " 1";
						++pc;
					}
					Arrays.sort(c);
					lines.add(pc+ans);
					continue;
				}
				int nind = -1;
				for (int j = 0; j < n; ++j) {
					if (c[j]==ct)
						nind = j;
				}
				int mid = n-nind-ct;
				if (nind != 0) {
					ans += " "+nind;
					pc++;
				}
				ans += " "+mid;
				pc++;
				for (int j = 0; j < ct; ++j) {
					pc++;
					ans += " 1";
				}
			} else {
				end = !end;
				for (int j = 0; j < n; ++j) {
					if (j==c[j])
						++ct;
					else
						break;
				}
				int nind = -1;
				for (int j = 0; j < n; ++j) {
					if (c[j]==ct)
						nind = j;
				}
				int mid = nind-ct+1;
				for (int j = 0; j < ct; ++j) {
					pc++;
					ans += " 1";
				}
				ans += " "+mid;
				pc++;
				if (nind != n-1) {
					ans += " "+(n-nind-1);
					pc++;
				}
			}
			if (pc==1)
				continue;
			lines.add(pc+ans);
			int[] piles = Arrays.stream(ans.split(" ")).filter(q->!q.isEmpty()).mapToInt(q->Integer.parseInt(q)).toArray();
			int[] nc = new int[n];
			int a = 0;
			int b = n-1;
			for (int ab : piles) {
				int ta = a+ab-1;
				for (int j = 0; j < ab; ++j) {
					nc[b] = c[ta];
					--b;
					--ta;
				}
				a += ab;
			}
			c = nc;
		}
		System.out.println(lines.size());
		for (String l : lines)
			System.out.println(l);
	}
}