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.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.stream.IntStream;
import java.util.Arrays;
import java.util.List;
import java.io.IOException;
import java.io.InputStream;
import java.util.ArrayList;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 */
public class Main {
    public static void main(String[] args) throws Exception {
        Thread thread = new Thread(null, new TaskAdapter(), "", 1 << 29);
        thread.start();
        thread.join();
    }

    static class TaskAdapter implements Runnable {
        @Override
        public void run() {
            InputStream inputStream = System.in;
            OutputStream outputStream = System.out;
            FastInput in = new FastInput(inputStream);
            PrintWriter out = new PrintWriter(outputStream);
            DUnshufflingADeck solver = new DUnshufflingADeck();
            solver.solve(1, in, out);
            out.close();
        }
    }

    static class DUnshufflingADeck {
        int[] c;
        List<int[]> op = new ArrayList<>();

        public void solve(int testNumber, FastInput in, PrintWriter out) {
            int n = in.readInt();
            c = new int[n];

            in.populate(c);
            int[] sorted = c.clone();
            Arrays.sort(sorted);

            int curL = n / 2;
            int curR = curL;
            int first = indexOf(0, n - 1, sorted[curL]);
            boolean head;

            apply(first, n - first);
            head = true;
            for (int i = 1; i < n; i++) {
                int l = 0, r = n - 1;
                int target;
                if (head) {
                    l += i;
                    target = sorted[--curL];
                } else {
                    r -= i;
                    target = sorted[++curR];
                }
                int j = indexOf(l, r, target);
                if (head) {
                    apply(i, j - i + 1, n - i - (j - i + 1));
                } else {
                    apply(j, n - j - i, i);
                }
                head = !head;
            }

            assert CompareUtils.notStrictAscending(c, 0, n - 1);
            out.println(op.size());
            for (int[] x : op) {
                out.print(x.length);
                out.append(' ');
                for (int y : x) {
                    out.print(y);
                    out.append(' ');
                }
                out.println();
            }
        }

        public int indexOf(int l, int r, int x) {
            for (int j = l; j <= r; j++) {
                if (c[j] == x) {
                    return j;
                }
            }
            throw new RuntimeException();
        }

        public int[] removeZero(int[] d) {
            int wpos = 0;
            for (int i = 0; i < d.length; i++) {
                if (d[i] > 0) {
                    d[wpos++] = d[i];
                }
            }
            return Arrays.copyOf(d, wpos);
        }

        public void apply(int... d) {
            d = removeZero(d);
            if (d.length <= 1) {
                return;
            }
            assert Arrays.stream(d).sum() == c.length;
            int[] next = new int[c.length];
            int wpos = 0;
            int r = c.length - 1;
            for (int i = d.length - 1; i >= 0; i--) {
                int len = d[i];
                int l = r - len + 1;
                for (int j = l; j <= r; j++) {
                    next[wpos++] = c[j];
                }
                r = l - 1;
            }

            c = next;
            op.add(d);
        }

    }

    static class CompareUtils {
        private CompareUtils() {
        }

        public static boolean notStrictAscending(int[] data, int l, int r) {
            for (int i = l + 1; i <= r; i++) {
                if (data[i] < data[i - 1]) {
                    return false;
                }
            }
            return true;
        }

    }

    static class FastInput {
        private final InputStream is;
        private byte[] buf = new byte[1 << 13];
        private int bufLen;
        private int bufOffset;
        private int next;

        public FastInput(InputStream is) {
            this.is = is;
        }

        public void populate(int[] data) {
            for (int i = 0; i < data.length; i++) {
                data[i] = readInt();
            }
        }

        private int read() {
            while (bufLen == bufOffset) {
                bufOffset = 0;
                try {
                    bufLen = is.read(buf);
                } catch (IOException e) {
                    bufLen = -1;
                }
                if (bufLen == -1) {
                    return -1;
                }
            }
            return buf[bufOffset++];
        }

        public void skipBlank() {
            while (next >= 0 && next <= 32) {
                next = read();
            }
        }

        public int readInt() {
            int sign = 1;

            skipBlank();
            if (next == '+' || next == '-') {
                sign = next == '+' ? 1 : -1;
                next = read();
            }

            int val = 0;
            if (sign == 1) {
                while (next >= '0' && next <= '9') {
                    val = val * 10 + next - '0';
                    next = read();
                }
            } else {
                while (next >= '0' && next <= '9') {
                    val = val * 10 - next + '0';
                    next = read();
                }
            }

            return val;
        }

    }
}