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.util.*;

public class Main {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int[] ta = new int[n];
        for (int i = 0; i < n; i ++)
            ta[i] = sc.nextInt();
        int res = 0;
        StringBuffer sb = new StringBuffer();
        while (true) {
            Map<Integer, Integer> map = new HashMap();
            for (int i = 0; i < n; i ++) {
                map.put(ta[i], i);
            }
            boolean ok = false;
            for (int i = 0; i < n; i ++) {
                int val = ta[i];
                int ti = map.getOrDefault(val - 1, -1);
                int cnt = 0;
                int tl = 1;
                if (ti > i) {
                    //System.out.println("i : " + i + " "  + "ti : " + ti);
                    StringBuffer tsb = new StringBuffer();
                    res++;
                    ok = true;
                    int tii = ti - 1;
                    while (tii >= 0 && ta[tii] + 1 == ta[tii + 1]) {
                        tl++;
                        tii--;
                    }
                    //System.out.println("tl : " + tl);
                    int[] nta = new int[n];
                    if (i > 0) {
                        tsb.append(i + " ");
                        for (int j = 0; j < i; j ++) {
                            nta[n - i + j] = ta[j];
                        }
                        cnt++;
                    }
                    //System.out.println("12");
                    //2
                    cnt++;
                    int tll = (ti - tl + 1 - i);
                    tsb.append(tll + " ");
                    //System.out.println("tll : " + tll);
                    for (int j = i; j < ti - tl + 1; j ++) {
                        nta[n - i - tll + j - i] = ta[j];
                    }
                    //3
                    //System.out.println("13");
                    cnt++;
                    tsb.append(tl + " ");
                    for (int j = ti - tl + 1; j <= ti; j ++) {
                        nta[n - i - tll - tl + j - (ti - tl + 1)] = ta[j];
                    }
                    //System.out.println("14");
                    if (ti != n - 1) {
                        tsb.append((n - 1 - ti));
                        cnt++;
                        for (int j = ti + 1; j < n; j ++) {
                            nta[j - (ti + 1)] = ta[j];
                        }
                    }
                    //System.out.println("15");
                    sb.append(cnt + " " + tsb.toString() + "\n");
                    ta = nta;

                    break;
                }

            }

//            for (int val : ta)
//                System.out.print(val + " ");
//            System.out.println();
            if (!ok) break;
        }
        System.out.println(res);
        System.out.println(sb.toString());
    }
}

/*

import java.util.*;

public class Main {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int T = sc.nextInt();

    }
}


 */