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.io.BufferedOutputStream;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.List;
import java.util.StringTokenizer;
import java.util.stream.Collectors;
import java.util.stream.IntStream;

public class D {
    private static class Solver {
        private void solve() throws Exception {
            int n = in.nextInt();
            Integer[] cards = new Integer[n + 1];
            for (int i = 1; i <= n; i++) {
                cards[i] = in.nextInt();
            }
            List<List<Integer>> ops = reorder(n, cards, false);
            if (ops.size() > n) {
                ops = reorder(n, cards, true);
            }
            out.println(ops.size());
            for (List<Integer> p : ops) {
                out.printf("%d %s\n", p.size(), p.stream().map(String::valueOf)
                        .collect(Collectors.joining(" ")));
            }
        }

        private List<List<Integer>> reorder(int n, Integer[] cards, boolean reverse) {
            List<List<Integer>> ops = new ArrayList<>();
            if (reverse) {
                Integer[] op = new Integer[n];
                Arrays.fill(op, 1);
                ops.add(Arrays.asList(op));
                for (int i = n; i > n - i + 1; i--) {
                    int temp = cards[n - i + 1];
                    cards[n - i + 1] = cards[i];
                    cards[i] = temp;
                }
            }
            while (true) {
                boolean inorder = true;
                boolean last = true;
                int target = cards[1] == 1 ? n : cards[1] - 1;
                int maxPrefix = 1;
                int targetIndex = 0;
                int beginIndex = 0;
                for (int i = 1; i <= n; i++) {
                    if (i != cards[i]) {
                        inorder = false;
                    }
                    if (i > 1 && cards[i] == cards[i - 1] + 1 && maxPrefix == i - 1) {
                        maxPrefix = i;
                    }
                    if (i > 1 && cards[i - 1] % n + 1 != cards[i]) {
                        last = false;
                    }
                    if (cards[i] == 1) {
                        beginIndex = i;
                    }
                    if (cards[i] == target) {
                        targetIndex = i;
                    }
                }
                if (inorder) {
                    break;
                }
                if (last) {
                    ops.add(Arrays.asList(beginIndex - 1, n - beginIndex + 1));
                    break;
                }
                List<Integer> op = new ArrayList<>();
                op.add(maxPrefix);
                op.add(targetIndex - maxPrefix);
                if (n > targetIndex) {
                    op.add(n - targetIndex);
                }
                ops.add(op);
                int newIndex = 1;
                Integer[] newCards = new Integer[n + 1];
                for (int i = targetIndex + 1; i <= n; i++) {
                    newCards[newIndex++] = cards[i];
                }
                for (int i = maxPrefix + 1; i <= targetIndex; i++) {
                    newCards[newIndex++] = cards[i];
                }
                for (int i = 1; i <= maxPrefix; i++) {
                    newCards[newIndex++] = cards[i];
                }
                cards = newCards;
                if (ops.size() > n) {
                    return ops;
                }
            }
            return ops;
        }
    }

    //--------------------------------------------------------
    private static final MyScanner in = new MyScanner(System.in);
    private static final PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));

    public static void main(String[] args) throws Exception {
        Solver solver = new Solver();
        solver.solve();
        out.close();
    }

    public static class MyScanner {
        private final BufferedReader br;
        private StringTokenizer st;

        public MyScanner(InputStream in) {
            br = new BufferedReader(new InputStreamReader(in));
        }

        String next() throws IOException {
            while (st == null || !st.hasMoreElements()) {
                st = new StringTokenizer(br.readLine());
            }
            return st.nextToken();
        }

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

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

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

        String nextLine() throws IOException {
            return br.readLine();
        }
    }
}