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.*;

public class D {
    static int n;
    static int[] arr;

    public static void main(String[] args) throws IOException {
        Scanner sc = new Scanner(System.in);
        n = sc.nextInt();
        arr = sc.nextIntArr(n);
        ArrayList<Integer>[] ops = new ArrayList[n];
        int count;
        for (count = 0; count < n; count++) {
            if (sorted(arr))
                break;
            boolean[] visited = new boolean[n+1];
            visited[0] = true;
            for (int i = 0; i < n; i++) {
                visited[arr[i]] = true;
                if (!visited[arr[i]-1]) {
                    int j;
                    for (j = n-1; j > i; j--) {
                        if (arr[j] == arr[i]-1)
                            break;
                    }
                    int t = j;
                    while (t > 0 && arr[t-1] == arr[t]-1)
                        t--;
                    ops[count] = operate(i, j, t);
                    break;
                }
            }
        }
        StringBuffer sb = new StringBuffer();
        sb.append(count + "\n");
        for (int i = 0; i < count; i++) {
            sb.append(ops[i].size() + " ");
            for (int j : ops[i])
                sb.append(j + " ");
            sb.append("\n");
        }
        System.out.print(sb.toString());
    }

    static ArrayList<Integer> operate(int i, int j, int t) {
        ArrayList<Integer> op = new ArrayList<Integer>();
        int mid1 = t-i;
        int mid2 = j-t+1;
        int end = n-j-1;
        if (i > 0)
            op.add(i);
        op.add(mid1);
        op.add(mid2);
        if (end > 0)
            op.add(end);
        int[] next = new int[n];
        Collections.reverse(op);
        int right = n-1;
        int x = 0;
        for (int len : op) {
            int y = right-len+1;
            while (y <= right)
               next[x++] = arr[y++];
            right -= len;
        }
        for (int k = 0; k < n; k++)
            arr[k] = next[k];
        Collections.reverse(op);
        return op;
    }

    static boolean sorted(int[] arr) {
        for (int i=0; i < arr.length; i++)
            if (arr[i] != i+1) return false;
        return true;
    }

    static class Scanner {
        StringTokenizer st;
        BufferedReader br;

        public Scanner(FileReader r) {
            br = new BufferedReader(r);
        }

        public Scanner(InputStream s) {
            br = new BufferedReader(new InputStreamReader(s));
        }

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

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

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

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

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

        public double nextDouble() throws IOException {
            String x = next();
            StringBuilder sb = new StringBuilder("0");
            double res = 0, f = 1;
            boolean dec = false, neg = false;
            int start = 0;
            if (x.charAt(0) == '-') {
                neg = true;
                start++;
            }
            for (int i = start; i < x.length(); i++)
                if (x.charAt(i) == '.') {
                    res = Long.parseLong(sb.toString());
                    sb = new StringBuilder("0");
                    dec = true;
                } else {
                    sb.append(x.charAt(i));
                    if (dec)
                        f *= 10;
                }
            res += Long.parseLong(sb.toString()) / f;
            return res * (neg ? -1 : 1);
        }

        public boolean ready() throws IOException {
            return br.ready();
        }
    }
}