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.*;
import java.util.*;

public class Main {
//    static ArrayList<ArrayList<Integer>> permutations = new ArrayList<>();
    public static void main(String[] args) {
        FastScanner in = new FastScanner();
        int n = in.nextInt();
        int[] a = in.readArray(n);
        for (int i = 0; i < n; i++) a[i]--;
        int[] temp = new int[n];
        int steps = 0;
        ArrayList<ArrayList<Integer>> ans = new ArrayList<>();
        while (true) {
            steps++;
            int idx;

            for (idx = n - 1; idx >= 0; idx--) {
                if (a[idx] != idx) break;
            }
            if (idx == -1) break;
            // find a[idx] + 1
            int pos = -1;
            for (int i = 0; i < idx; i++) {
                if (a[i] == a[idx] + 1) {
                    pos = i;
                }
            }
            int st = idx;
            int v = a[idx];
            while (a[st] == v) {
                st--;
                v--;
            }

            int[] l = new int[]{0, pos, st + 1, idx + 1};
            int[] r = new int[]{pos - 1, st, idx, n - 1};
            ArrayList<Integer> lo = new ArrayList<>();
            ArrayList<Integer> hi = new ArrayList<>();
            ArrayList<Integer> now = new ArrayList<>();
            for (int i = 0; i < 4; i++) {
                if (l[i] <= r[i]) {
                    lo.add(l[i]);
                    hi.add(r[i]);
                    now.add(r[i] - l[i] + 1);
                }
            }
            ans.add(now);
            int ptr = 0;
            for (int i = lo.size() - 1; i >= 0; i--) {
                for (int j = lo.get(i); j <= hi.get(i); j++) {
                    temp[ptr++] = a[j];
                }
            }
            for(int i = 0; i < n; i++) a[i] = temp[i];
        }
        System.out.println(ans.size());
        for(int i = 0; i < ans.size(); i++){
            System.out.print(ans.get(i).size() + " ");
            for(int e : ans.get(i)){
                System.out.print(e + " ");
            }
            System.out.println();
        }
    }

//    static void generatePerm(int i, ArrayList<Integer> a) {
//        if(i != -1) { a.add(i); }
//        if (a.size() == 7){
//            permutations.add(a);
//            return;
//        }
//        for(int j = 0; j < 7; j ++) {
//            if(!a.contains(j)){
//                ArrayList<Integer> b = new ArrayList<>(a);
//                generatePerm(j, b);
//            }
//        }
//    }

    // [3  12 4]
    // [4 123]
    // [1234]

    // [3 512 4]
    // [4512 3]
    // []

    // [4,3,2,1]
    // [1]


    // [6 3 4 1 5 2]
    // [6 3415 2]
    // [234156]
    // [234 1 56]
    // [56 1234]
    // [123456]
    static class FastScanner {
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st=new StringTokenizer("");
        String next() {
            while (!st.hasMoreTokens())
                try {
                    st=new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            return st.nextToken();
        }

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

        int[] readArray(int n) {
            int[] a=new int[n];
            for (int i=0; i<n; i++) a[i]=nextInt();
            return a;
        }
        long nextLong() {
            return Long.parseLong(next());
        }
    }

}
