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 javax.print.DocFlavor;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.nio.file.Paths;
import java.util.*;


public class Main extends PrintWriter {
    static BufferedReader s = new BufferedReader(new InputStreamReader(System.in));
    Main () { super(System.out); }
    public static void main(String[] args) throws IOException{
        Main  d1=new Main   ();d1.main();d1.flush();
    }
    void main() throws IOException {
        BufferedReader s = new BufferedReader(new InputStreamReader(System.in));
        StringBuilder sb = new StringBuilder();
        PrintWriter out = new PrintWriter(System.out);
        int t=1;
     //   t=i(s()[0]);
        while(t-->0) {
            String[] s1=s();
            int n=i(s1[0]);
            int[] a=new int[n];int cnt=0;arri(a,n);
            while(true){
                HashMap<Integer,Integer> h=new HashMap<>();
                int flag=0;
                int ind1=0;int ind2=0;
                for(int i=0;i<n;i++){
                    if(h.containsKey(a[i]+1)){
                        flag=1;ind2=i;ind1=h.get(a[i]+1);break;
                    }
                    h.put(a[i],i);
                }int j=0;
                if(flag==0) break;
                cnt++;
                int beech=ind1;
                for(int i=beech+1;i<ind2;i++){
                    if(a[i]!=a[i-1]+1){
                        beech=i-1;break;
                    }
                }
                if(ind1==0&&n-ind2-1==0){
                    sb.append(2+" "+(beech-ind1+1)+" "+(n-1-beech)+"\n");
                }else if(ind1==0){
                    sb.append(3+" "+(beech-ind1+1)+" "+(ind2-beech)+" "+(n-ind2-1)+"\n");
                }else if(n-ind2-1==0){
                    sb.append(3+" "+ind1+" "+(beech-ind1+1)+" "+(ind2-beech)+"\n");
                }else
                  sb.append(4+" "+(ind1+" "+(beech-ind1+1)+" "+(ind2-beech)+" "+(n-ind2-1)+"\n"));
                int[] b=new int[n];
                for(int i=ind2+1;i<n;i++){
                    b[j++]=a[i];
                }
                for(int i=beech+1;i<=ind2;i++){
                    b[j++]=a[i];
                }
                for(int i=ind1;i<=beech;i++){
                    b[j++]=a[i];
                }
                for(int i=0;i<ind1;i++){
                    b[j++]=a[i];
                }
                for(int i=0;i<n;i++) {a[i]=b[i];
              // System.out.print(a[i]+" ");
                }
             //  System.out.println();
                }
            System.out.println(cnt);
            System.out.println(sb);
        }
    }TreeMap<Long,Integer> tm;
    int[] pathCompression(long[] a){
        int n=a.length;
        int[] compressed=new int[n];
        class Pair1{ long orgVal;int indexBeforeSorting; public Pair1(long a,int b){orgVal=a;indexBeforeSorting=b;}}
        Pair1[] p=new Pair1[n];
        for(int i=0;i<n;i++){
            p[i]=new Pair1(a[i],i);
        }int pre=0;
        Arrays.sort(p, new Comparator<Pair1>() {@Override public int compare(Pair1 o1, Pair1 o2) { if(o1.orgVal<o2.orgVal) return -1; else if(o1.orgVal==o2.orgVal) return 0; return 1; }});
        for(int i=0;i<p.length;i++){if(i==0||p[i].orgVal==p[i-1].orgVal) compressed[p[i].indexBeforeSorting]=pre; else {compressed[p[i].indexBeforeSorting]=++pre;}

        }
        return compressed;
    }
    long[] st;//new long[4*n];
    long query(int i,int s,int e,int qs,int qe){
        if(qs>e||qe<s) return Integer.MAX_VALUE;
        if(s>=qs&&e<=qe) return st[i];
        int mid=(s+e)/2;
        long l=query(2*i,s,mid,qs,qe);
        long r=query(2*i+1,mid+1,e,qs,qe);
        return Math.min(l,r);
    }
    void pointupdate(int i,int s,int e,int qi,long  a){
        if(s==e){
            st[i]=(a);return;
        }
        int mid=(s+e)/2;
        if(qi<=mid) pointupdate(2*i,s,mid,qi,a);
        else pointupdate(2*(i)+1,mid+1,e,qi,a);
        st[i]=Math.min(st[2*i],st[2*i+1]);
    }
    static long gcd(long a, long b) {
        if (b == 0)
            return a;
        return gcd(b, a % b);
    }
     static String[] s() throws IOException {
        return s.readLine().trim().split("\\s+");
    }
    static int i(String ss) {
        return Integer.parseInt(ss);
    }
    static long l(String ss) {
        return Long.parseLong(ss);
    }
    public void arr(long[] a,int n) throws IOException {
        //        BufferedReader s = new BufferedReader(new InputStreamReader(System.in));
        String[] s2=s();
        for(int i=0;i<n;i++){
            a[i]=l(s2[i]);
        }
    }
    public void arri(int[] a,int n) throws IOException{
        //        BufferedReader s = new BufferedReader(new InputStreamReader(System.in));
        String[] s2=s();
        for(int i=0;i<n;i++){
            a[i]=i(s2[i]);
        }
    }
    public void sort(int[] a,int l,int h){
        if(l==h) return;
        int mid=(l+h)/2;

        sort(a,l,mid);
        sort(a,mid+1,h);
        merge(a,l,(l+h)/2,h);
    }
    void merge(int arr[], int l, int m, int r) {
        int n1 = m - l + 1;
        int n2 = r - m;
        int L[] = new int[n1];
        int R[] = new int[n2];
        for (int i = 0; i < n1; ++i)
            L[i] = arr[l + i];
        for (int j = 0; j < n2; ++j)
            R[j] = arr[m + 1 + j];
        int i = 0, j = 0;
        int k = l;
        while (i < n1 && j < n2) {
            if (L[i] <= R[j]) {
                arr[k] = L[i];
                i++;
            }
            else {
                arr[k] = R[j];
                j++;
            }
            k++;
        }

        while (i < n1) {
            arr[k] = L[i];
            i++;
            k++;
        }
        while (j < n2) {
            arr[k] = R[j];
            j++;
            k++;
        }
    }
}
class Pair {
    long a;
    int b,c,d;
    public Pair(long a,int b){
        this.a=a;this.b=b;
        }
}class Sortbyroll implements Comparator<Pair> {
    public int compare(Pair a, Pair b)
    {
      return a.b-b.b;

    }}
