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 CF1427D extends PrintWriter {
	CF1427D() { super(System.out); }
	Scanner sc = new Scanner(System.in);
	public static void main(String[] $) {
		CF1427D o = new CF1427D(); o.main(); o.flush();
	}

	int[] aa, bb;
	void move(int m, int n) {
		System.arraycopy(aa, 0, bb, n, m);
		System.arraycopy(aa, m, bb, 0, n);
		System.arraycopy(bb, 0, aa, 0, m + n);
	}
	void main() {
		int n = sc.nextInt();
		aa = new int[n];
		bb = new int[n];
		for (int i = 0; i < n; i++)
			aa[i] = sc.nextInt() - 1;
		int[] ii = new int[n];
		int[][] op = new int[n - 1][]; int cnt = 0;
		while (true) {
			for (int i = 0; i < n; i++) {
				int a = aa[i];
				ii[a] = i;
			}
			int a = 1;
			while (a < n && ii[a - 1] < ii[a])
				a++;
			if (a == n)
				break;
			int i = ii[a], j = ii[a - 1], k = 2;
			if (i > 0)
				k++;
			if (j < n - 1)
				k++;
			int[] dd = op[cnt++] = new int[k]; k = 0;
			if (i > 0)
				dd[k++] = i;
			int d = 1;
			while (aa[i + d] == aa[i] + d)
				d++;
			dd[k++] = d;
			dd[k++] = j - i + 1 - d;
			if (j < n - 1)
				dd[k++] = n - 1 - j;
			d = dd[0];
			for (int h = 1; h < k; h++) {
				move(d, dd[h]);
				d += dd[h];
			}
		}
		println(cnt);
		for (int o = 0; o < cnt; o++) {
			int[] dd = op[o];
			int k = dd.length;
			print(k);
			for (int h = 0; h < k; h++)
				print(" " + dd[h]);
			println();
		}
	}
}