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.*;
import java.math.BigInteger;

public class C {
	FastScanner in;
	PrintWriter out;
	boolean systemIO = true;

	public static void quickSort(int[] a, int from, int to) {
		if (to - from <= 1) {
			return;
		}
		int i = from;
		int j = to - 1;
		int x = a[from + (new Random()).nextInt(to - from)];
		while (i <= j) {
			while (a[i] < x) {
				i++;
			}
			while (a[j] > x) {
				j--;
			}
			if (i <= j) {
				int t = a[i];
				a[i] = a[j];
				a[j] = t;
				i++;
				j--;
			}
		}
		quickSort(a, from, j + 1);
		quickSort(a, j + 1, to);
	}

	public long gcd(long x, long y) {
		if (y == 0) {
			return x;
		}
		if (x == 0) {
			return y;
		}
		return gcd(y, x % y);
	}

	public boolean prime(long x) {
		for (int i = 2; i * i <= x; i++) {
			if (x % i == 0) {
				return false;
			}
		}
		return true;
	}

	public long pow(long x, long p) {
		if (p == 0) {
			return 1;
		}
		long t = pow(x, p / 2);
		t *= t;
		t %= mod;
		if (p % 2 == 1) {
			t *= x;
			t %= mod;
		}
		return t;
	}

	public class Pair implements Comparable<Pair> {
		int x;
		int y;

		public Pair(int x, int y) {
			this.x = x;
			this.y = y;
		}
		
		public String toString() {
			return x + " " + y;
		}

		@Override
		public int compareTo(Pair o) {
			if (y > o.y) {
				return 1;
			}
			if (y < o.y) {
				return -1;
			}
			if (x > o.x) {
				return 1;
			}
			if (x < o.x) {
				return -1;
			}
			return 0;
		}
	}

	long mod = 1000000;

	public class Fenvik2D {
		long[][] t;
		int n, m;
		
		public Fenvik2D(int n, int m) {
			t = new long[n][m];
			this.n = n;
			this.m = m;
		}

		public int sum(int x1, int y1, int x2, int y2) {
			return sum(x2, y2) - sum(x1 - 1, y2) - sum(x2, y1 - 1) + sum(x1 - 1, y1 - 1);
		}
		
		public int sum(int x, int y) {
			int result = 0;
			for (int i = x; i >= 0; i = (i & (i+1)) - 1) {
				for (int j = y; j >= 0; j = (j & (j+1)) - 1) {
					result += t[i][j];
				}
			}
			return result;
		}

		public void add(int x, int y, int delta) {
			for (int i = x; i < n; i = (i | (i+1))) {
				for (int j = y; j < m; j = (j | (j+1))) {
					t[i][j] += delta;
				}
			}
		}
	}
	
	public int[] prefF(int[] a) {
		int[] p = new int[a.length];
		for (int i = 1; i < p.length; i++) {
			int j = p[i - 1];
			while (j > 0) {
				if (a[j] == a[i]) {
					break;
				}
				if (j == 0) {
					break;
				}
				j = p[j - 1];
			}
			if (a[i] == a[j]) {
				p[i] = j + 1;
			}
		}
		return p;
	}

	public class Tree {
		long[] sum;
		long[] min;
		int[] alive;
		int pow;
		long inf = Long.MAX_VALUE / 2;
		
		public Tree(long[] a, long[] b) {
			pow = 1;
			while (pow < a.length) {
				pow *= 2;
			}
			sum = new long[2 * pow];
			min = new long[2 * pow];
			for (int i = 0; i < min.length; i++) {
				min[i] = inf;
			}
			alive = new int[2 * pow];
			for (int i = 0; i < b.length; i++) {
				sum[pow + i] = a[i];
				min[pow + i] = a[i] + b[i];
				alive[pow + i] = 1;
			}
			for (int i = pow - 1; i > 0; i--) {
				merge(i);
			}
		}
		
		public void merge(int v) {
			sum[v] = sum[2 * v] + sum[2 * v + 1];
			min[v] = inf;
			if (min[2 * v] < min[v]) {
				min[v] = min[2 * v];
			}
			if (min[2 * v + 1] + sum[2 * v] < min[v]) {
				min[v] = min[2 * v + 1] + sum[2 * v];
			}
			alive[v] = alive[2 * v] + alive[2 * v + 1];
		}
		
		public void add(long a, long b, int c) {
			add(1, 0, pow, a, b, c - 1);
		}
		
		private void add(int v, int l, int r, long a, long b, int c) {
			if (v >= pow) {
				sum[v] = a;
				min[v] = a + b;
				alive[v] = 1;
				return;
			}
			int m = (l + r) / 2;
			if (c < m) {
				add(2 * v, l, m, a, b, c);
			} else {
				add(2 * v + 1, m, r, a, b, c);
			}
			merge(v);
		}
		
		public int update(long x) {
			return update(1, 0, pow, x);
		}
		
		private int update(int v, int l, int r, long x) {
			if (v >= pow) {
				if (x > min[v]) {
					sum[v] = 0;
					min[v] = inf;
					alive[v] = 0;
					return 1;
				}
				if (alive[v] == 1 && x >= sum[v]) {
					return 1;
				}
				return 0;
			}
			int m = (l + r) / 2;
			if (sum[2 * v] > x) {
				int ans = update(2 * v, l, m, x);
				merge(v);
				return ans;
			}
			int ans = alive[2 * v] + update(2 * v + 1, m, r, x - sum[2 * v]);
			if (x > min[2 * v]) {
				update(2 * v, l, m, x);
			}
			merge(v);
			return ans;
		}
	}
	
	public class Stack {
		int[] st;
		int sz;
		
		public Stack(int max) {
			st = new int[max];
			sz = 0;
		}
		
		public int size() {
			return sz;
		}
		
		public boolean isEmpty() {
			return sz == 0;
		}
		
		public int peek() {
			return st[sz - 1];
		}
		
		public int pop() {
			return st[--sz];
		}
		
		public void add(int x) {
			st[sz++] = x;
		}
		
		public void clear() {
			sz = 0;
		}
	}

	public long[][] mult(long[][] a, long[][] b) {
		long[][] c = new long[a.length][b[0].length];
		for (int i = 0; i < c.length; i++) {
			for (int j = 0; j < c[0].length; j++) {
				for (int k = 0; k < b.length; k++) {
					c[i][j] += a[i][k] * b[k][j];
					c[i][j] %= mod;
				}
			}
		}
		return c;
	}
	
	public long[][] pow(long p, long[][] m) {
		if (p == 0) {
			long[][] matrix = new long[m.length][m.length];
			for (int i = 0; i < matrix.length; i++) {
				matrix[i][i] = 1;
			}
			return matrix;
		}
		if (p % 2 == 1) {
			return mult(m, pow(p - 1, m));
		}
		long[][] cur = pow(p / 2, m);
		return mult(cur, cur);
	}
	
	public void add(HashMap<Integer, Integer> map, int x) {
		if (map.containsKey(x)) {
			map.put(x, map.get(x) + 1);
		} else {
			map.put(x, 1);
		}
	}
	
	public int[] permut(int[] a, ArrayList<Integer> list) {
		int[] ans = new int[a.length];
		int last = a.length;
		int c = 0;
		for (int i = list.size() - 1; i >= 0; i--) {
			for (int j = last - list.get(i); j < last; j++) {
				ans[c++] = a[j];
			}
			last -= list.get(i);
		}
		return ans;
	}
	
	public boolean identical(int[] perm) {
		for (int i = 0; i < perm.length; i++) {
			if (perm[i] != i) {
				return false;
			}
		}
		return true;
	}
	
	public void solve() {
		long time = System.currentTimeMillis();
		int n = in.nextInt();
		int[] c = new int[n];
		for (int i = 0; i < c.length; i++) {
			c[i] = in.nextInt() - 1;
		}
		ArrayList<ArrayList<Integer>> ans = new ArrayList<>();
		for (int i = 0; !identical(c); i++) {
			int cur = i / 2;
			if (i % 2 == 1) {
				cur = n - 1 - cur;
			}
			ArrayList<Integer> list = new ArrayList<>();
			int place = -1;
			for (int j = 0; j < c.length; j++) {
				if (c[j] == cur) {
					place = j;
				}
			}
			int pref = i / 2 + i % 2;
			int suf = i / 2;
			for (int j = 0; j < pref; j++) {
				list.add(1);
			}
			if (place - pref > 0) {
				list.add(place - pref);
			}
			list.add(n - place - suf);
			for (int j = 0; j < suf; j++) {
				list.add(1);
			}
			c = permut(c, list);
			if (list.size() > 1) {
				ans.add(list);
			}
		}
		out.println(ans.size());
		for (ArrayList<Integer> list : ans) {
			out.print(list.size() + " ");
			for (int i : list) {
				out.print(i + " ");
			}
			out.println();
		}
		System.err.println(System.currentTimeMillis() - time);
	}

	public void run() {
		try {
			if (systemIO) {
				in = new FastScanner(System.in);
				out = new PrintWriter(System.out);
			} else {
				in = new FastScanner(new File("input.txt"));
				out = new PrintWriter(new File("output.txt"));
			}
			solve();

			out.close();
		} catch (IOException e) {
			e.printStackTrace();
		}
	}

	class FastScanner {
		BufferedReader br;
		StringTokenizer st;

		FastScanner(File f) {
			try {
				br = new BufferedReader(new FileReader(f));
			} catch (FileNotFoundException e) {
				e.printStackTrace();
			}
		}

		FastScanner(InputStream f) {
			br = new BufferedReader(new InputStreamReader(f));
		}

		String nextLine() {
			try {
				return br.readLine();
			} catch (IOException e) {
				return null;
			}
		}

		String next() {
			while (st == null || !st.hasMoreTokens()) {
				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());
		}

	}

	// AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
	public static void main(String[] arg) {
		new C().run();
	}
}