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 static java.lang.Integer.parseInt;
import static java.lang.Long.parseLong;
import static java.lang.System.exit;

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.List;
import java.util.StringTokenizer;

public class D {

	static void solve() throws Exception {
		int n = scanInt();
		int a[] = new int[n], aInv[] = new int[n], na[] = new int[n];
		for (int i = 0; i < n; i++) {
			aInv[a[i] = scanInt() - 1] = i;
		}
		List<int[]> ans = new ArrayList<>();
		find: while (true) {
			int ci;
			for (ci = 0;; ci++) {
				if (ci >= n - 1) {
					break find;
				}
				if (aInv[ci + 1] < aInv[ci]) {
					break;
				}
			}
			int p1 = aInv[ci + 1], p3 = aInv[ci] + 1, p2;
			for (p2 = p1 + 1; a[p2 - 1] + 1 == a[p2]; p2++) { }
			int len = 2 + (p1 != 0 ? 1 : 0) + (p3 != n ? 1 : 0);
			int cans[] = new int[len], cpos = 0;
			if (p1 != 0) {
				cans[cpos++] = p1;
			}
			cans[cpos++] = p2 - p1;
			cans[cpos++] = p3 - p2;
			if (p3 != n) {
				cans[cpos++] = n - p3;
			}
			ans.add(cans);
			for (int pOld = 0, pNew = n, j = 0; j < len; j++) {
				int clen = cans[j];
				pNew -= clen;
				for (int jj = 0; jj < clen; jj++) {
					aInv[na[pNew + jj] = a[pOld + jj]] = pNew + jj;
				}
				pOld += clen;
			}
			int t[] = a;
			a = na;
			na = t;
			if (ans.size() == n) {
				throw new AssertionError();
			}
		}
		out.println(ans.size());
		for (int cans[]: ans) {
			out.print(cans.length);
			for (int v: cans) {
				out.print(" " + v);
			}
			out.println();
		}
	}

	static int scanInt() throws IOException {
		return parseInt(scanString());
	}

	static long scanLong() throws IOException {
		return parseLong(scanString());
	}

	static String scanString() throws IOException {
		while (tok == null || !tok.hasMoreTokens()) {
			tok = new StringTokenizer(in.readLine());
		}
		return tok.nextToken();
	}

	static BufferedReader in;
	static PrintWriter out;
	static StringTokenizer tok;

	public static void main(String[] args) {
		try {
			in = new BufferedReader(new InputStreamReader(System.in));
			out = new PrintWriter(System.out);
			solve();
			in.close();
			out.close();
		} catch (Throwable e) {
			e.printStackTrace();
			exit(1);
		}
	}
}