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.ArrayList;
import java.util.Arrays;
import java.util.List;
import java.util.StringTokenizer;

public class D {
    private FastScanner in;
    private PrintWriter out;

    public D(FastScanner in, PrintWriter out) {
        this.in = in;
        this.out = out;
    }

    public static void main(String[] args) {
        try (
                FastScanner in = new FastScanner(System.in);
                PrintWriter out = new PrintWriter(System.out);
        ) {
            D solution = new D(in, out);
            solution.solve();
        } catch (IOException e) {
            e.printStackTrace();
        }
    }

    public void solve() throws IOException {
        int n = in.nextInt();
        int[] c = in.nextIntArray(n);
        List<Operation> result = new ArrayList<>(n);
        if (c[0] != 1) {
            int pos = getPos(1, c);
            List<Integer> opParts = new ArrayList<>();
            opParts.add(pos);
            opParts.add(n - pos);
            result.add(new Operation(opParts));
            c = doOperation(c, opParts);
        }
        boolean turn = true;
        for (int next = 2; next < n; next++) {
            int pos = getPos(next, c);
            List<Integer> opParts = new ArrayList<>();
            if (turn) {
                for (int i = 1; i < next; i++) {
                    opParts.add(1);
                }
                opParts.add(pos - next + 2);
                if (pos < n - 1) {
                    opParts.add(n - pos - 1);
                }
            } else {
                if(pos > 0){
                    opParts.add(pos);
                }
                opParts.add(n - pos - next + 1);
                for (int i = 1; i < next; i++) {
                    opParts.add(1);
                }
            }
            result.add(new Operation(opParts));
            c = doOperation(c, opParts);
            turn = !turn;
        }
        if (!turn) {
            List<Integer> opParts = new ArrayList<>(n);
            for (int i = 0; i < n; i++) {
                opParts.add(1);
            }
            result.add(new Operation(opParts));
        }
        out.println(result.size());
        for (Operation operation : result) {
            out.println(operation);
        }
    }

    int getPos(int val, int[] c) {
        for (int i = 0; i < c.length; i++) {
            if (val == c[i]) {
                return i;
            }
        }
        return -1;
    }

    int[] doOperation(int[] c, List<Integer> parts) {
        int[] res = new int[c.length];
        int startC = c.length;
        int startRes = 0;
        for (int i = parts.size() - 1; i >= 0; i--) {
            int size = parts.get(i);
            startC -= size;
            System.arraycopy(c, startC, res, startRes, size);
            startRes += size;
        }
        return res;
    }

    public static class FastScanner implements Closeable {
        BufferedReader bufferedReader;
        StringTokenizer stringTokenizer;

        public FastScanner(InputStream inputStream) {
            this.bufferedReader = new BufferedReader(new InputStreamReader(inputStream));
        }

        public FastScanner(File file) throws FileNotFoundException {
            this(new FileInputStream(file));
        }

        public String next() throws IOException {
            while (stringTokenizer == null || !stringTokenizer.hasMoreTokens()) {
                String line = bufferedReader.readLine();
                if (line == null) {
                    throw new EOFException("End of input stream is reached.");
                }
                stringTokenizer = new StringTokenizer(line);
            }
            return stringTokenizer.nextToken();
        }

        public String nextLine() throws IOException {
            stringTokenizer = null;
            return bufferedReader.readLine();
        }

        public int nextInt() throws IOException {
            return Integer.parseInt(next());
        }

        public long nextLong() throws IOException {
            return Long.parseLong(next());
        }

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

        public int[] nextIntArray(int n) throws IOException {
            int[] res = new int[n];
            for (int i = 0; i < n; i++) {
                res[i] = nextInt();
            }
            return res;
        }

        public List<Integer> nextIntList(int n) throws IOException {
            List<Integer> res = new ArrayList<>(n);
            for (int i = 0; i < n; i++) {
                res.add(nextInt());
            }
            return res;
        }

        @Override
        public void close() throws IOException {
            bufferedReader.close();
        }
    }

    private class Operation {
        List<Integer> parts;

        public Operation(List<Integer> parts) {
            this.parts = parts;
        }

        @Override
        public String toString() {
            StringBuilder stringBuilder = new StringBuilder();
            stringBuilder.append(parts.size());
            for (int part : parts) {
                stringBuilder.append(' ');
                stringBuilder.append(part);
            }
            return stringBuilder.toString();
        }
    }
}