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.util.ArrayList;
import java.util.PriorityQueue;

public class Main {

	public static void main(String[] args) {
		new Main();
	}

	public Main() {
		FastScanner fs = new FastScanner();
		java.io.PrintWriter out = new java.io.PrintWriter(System.out);
		solve(fs, out);
		out.flush();
	}

	public void solve(FastScanner fs, java.io.PrintWriter out) {
		int n = fs.nextInt();
		int[] c = new int[n];
		for (int i = 0;i < n;++ i) c[i] = fs.nextInt();
		ArrayList<ArrayList<Integer>> ans = new ArrayList<>();
		for (int i = 2;i <= n;) {
			int a = 0, b = 0;
			while(c[a] != i - 1) ++ a;
			while(c[b] != i) ++ b;
			if (a < b) {
				++ i;
			} else {
				ArrayList<Integer> query = new ArrayList<>();
				if (b > 0) query.add(b);
				int back = 1;
				while(c[a - back] == c[a - back + 1] - 1) ++ back;
				query.add(a - b - back + 1);
				query.add(back);
				if (a < n - 1) query.add(n - a - 1);
//				out.println(query);
				out.flush();
				{
					int[] c2 = new int[n];
					int last = 0;
					for (int j : query) {
						System.arraycopy(c, last, c2, n - last - j, j);
						last += j;
					}
					c = c2;
				}
				i = 2;
//				out.println(java.util.Arrays.toString(c));
				ans.add(query);
			}
		}
		out.println(ans.size());
		for (ArrayList<Integer> i : ans) {
			out.print(i.size());
			for (int j : i) out.print(" " + j);
			out.println();
		}
	}
}

class FastScanner {

	private final java.io.InputStream in = System.in;
	private final byte[] buffer = new byte[8192];
	private int ptr = 0;
	private int buflen = 0;

	private boolean hasNextByte() {
		if (ptr < buflen) return true;
		ptr = 0;
		try {
			buflen = in.read(buffer);
		} catch (java.io.IOException e) {
			e.printStackTrace();
		}
		return buflen > 0;
	}

	private byte readByte() {
		return hasNextByte() ? buffer[ptr++ ] : -1;
	}

	private static boolean isPrintableChar(byte c) {
		return 32 < c || c < 0;
	}

	private static boolean isNumber(int c) {
		return '0' <= c && c <= '9';
	}

	public boolean hasNext() {
		while (hasNextByte() && !isPrintableChar(buffer[ptr]))
			ptr++ ;
		return hasNextByte();
	}

	public String next() {
		if (!hasNext()) throw new java.util.NoSuchElementException();
		StringBuilder sb = new StringBuilder();
		byte b;
		while (isPrintableChar(b = readByte()))
			sb.appendCodePoint(b);
		return sb.toString();
	}

	public final char nextChar() {
		if (!hasNext()) throw new java.util.NoSuchElementException();
		return (char)readByte();
	}

	public final long nextLong() {
		if (!hasNext()) throw new java.util.NoSuchElementException();
		long n = 0;
		try {
			byte b = readByte();
			if (b == '-') {
				while (isNumber(b = readByte()))
					n = n * 10 + '0' - b;
				return n;
			} else if (!isNumber(b)) throw new NumberFormatException();
			do
				n = n * 10 + b - '0';
			while (isNumber(b = readByte()));
		} catch (java.util.NoSuchElementException e) {}
		return n;
	}

	public final int nextInt() {
		if (!hasNext()) throw new java.util.NoSuchElementException();
		int n = 0;
		try {
			byte b = readByte();
			if (b == '-') {
				while (isNumber(b = readByte()))
					n = n * 10 + '0' - b;
				return n;
			} else if (!isNumber(b)) throw new NumberFormatException();
			do
				n = n * 10 + b - '0';
			while (isNumber(b = readByte()));
		} catch (java.util.NoSuchElementException e) {}
		return n;
	}

	public double nextDouble() {
		return Double.parseDouble(next());
	}
}

class Arrays {

	public static void sort(final int[] array) {
		int l, min = 0xFFFFFFFF, max = 0;
		for (l = 0; l < array.length; ++l) {
			int i = array[l];
			min &= i;
			max |= i;
			if ((i & 0x80000000) == 0) break;
		}
		for (int r = l + 1; r < array.length; ++r) {
			int i = array[r];
			min &= i;
			max |= i;
			if ((i & 0x80000000) != 0) {
				array[r] = array[l];
				array[l++ ] = i;
			}
		}
		int use = min ^ max, bit = Integer.highestOneBit(use & 0x7FFFFFFF);
		if (bit == 0) return;
		sort(array, 0, l, use, bit);
		sort(array, l, array.length, use, bit);
	}

	private static void sort(final int[] array, final int left, final int right, final int use, int digit) {
		if (right - left <= 96) {
			for (int i = left + 1; i < right; ++i) {
				int tmp = array[i], tmp2, j;
				for (j = i; j > left && (tmp2 = array[j - 1]) > tmp; --j)
					array[j] = tmp2;
				array[j] = tmp;
			}
			return;
		}
		int l = left;
		while (l < right && (array[l] & digit) == 0)
			++l;
		for (int r = l + 1; r < right; ++r) {
			int i = array[r];
			if ((i & digit) == 0) {
				array[r] = array[l];
				array[l++ ] = i;
			}
		}
		if ((digit = Integer.highestOneBit(use & digit - 1)) == 0) return;
		sort(array, left, l, use, digit);
		sort(array, l, right, use, digit);
	}

	public static void sort(final long[] array) {
		int l;
		long min = 0xFFFFFFFFFFFFFFFFL, max = 0;
		for (l = 0; l < array.length; ++l) {
			long i = array[l];
			min &= i;
			max |= i;
			if ((i & 0x8000000000000000L) == 0) break;
		}
		for (int r = l + 1; r < array.length; ++r) {
			long i = array[r];
			min &= i;
			max |= i;
			if ((i & 0x8000000000000000L) != 0) {
				array[r] = array[l];
				array[l++ ] = i;
			}
		}
		long use = min ^ max, bit = Long.highestOneBit(use & 0x7FFFFFFFFFFFFFFFL);
		if (bit == 0) return;
		sort(array, 0, l, use, bit);
		sort(array, l, array.length, use, bit);
	}

	private static void sort(final long[] array, final int left, final int right, final long use, long digit) {
		if (right - left <= 96) {
			for (int i = left + 1, j; i < right; ++i) {
				long tmp = array[i], tmp2;
				for (j = i; j > left && (tmp2 = array[j - 1]) > tmp; --j)
					array[j] = tmp2;
				array[j] = tmp;
			}
			return;
		}
		int l = left;
		while (l < right && (array[l] & digit) == 0)
			++l;
		for (int r = l + 1; r < right; ++r) {
			long i = array[r];
			if ((i & digit) == 0) {
				array[r] = array[l];
				array[l++ ] = i;
			}
		}
		if ((digit = Long.highestOneBit(use & digit - 1)) == 0) return;
		sort(array, left, l, use, digit);
		sort(array, l, right, use, digit);
	}
}

class IntMath {

	public static int gcd(int a, int b) {
		while (a != 0)
			if ((b %= a) != 0) a %= b;
			else return a;
		return b;
	}

	public static int gcd(int... array) {
		int ret = array[0];
		for (int i = 1; i < array.length; ++i)
			ret = gcd(ret, array[i]);
		return ret;
	}

	public static long gcd(long a, long b) {
		while (a != 0)
			if ((b %= a) != 0) a %= b;
			else return a;
		return b;
	}

	public static long gcd(long... array) {
		long ret = array[0];
		for (int i = 1; i < array.length; ++i)
			ret = gcd(ret, array[i]);
		return ret;
	}

	public static long lcm(long a, long b) {
		return a / gcd(a, b) * b;
	}

	public static int pow(int a, int b) {
		int ans = 1;
		for (int mul = a; b > 0; b >>= 1, mul *= mul)
			if ((b & 1) != 0) ans *= mul;
		return ans;
	}

	public static long powLong(long a, long b) {
		long ans = 1;
		for (long mul = a; b > 0; b >>= 1, mul *= mul)
			if ((b & 1) != 0) ans *= mul;
		return ans;
	}

	public static int pow(int a, int b, int mod) {
		if (b < 0) b = b % (mod - 1) + mod - 1;
		long ans = 1;
		for (long mul = a; b > 0; b >>= 1, mul = mul * mul % mod)
			if ((b & 1) != 0) ans = ans * mul % mod;
		return (int)ans;
	}

	public static int floorsqrt(long n) {
		return (int)Math.sqrt(n + 0.1);
	}

	public static int ceilsqrt(long n) {
		return n <= 1 ? (int)n : (int)Math.sqrt(n - 0.1) + 1;
	}
}
