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.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.*;
import java.util.stream.Stream;

public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader in = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        solve(in, out);
        out.close();
    }

    static String reverse(String s) {
        return (new StringBuilder(s)).reverse().toString();
    }

    static void sieveOfEratosthenes(int n, int factors[]) 
    { 
        factors[1]=1;
        for(int p = 2; p*p <=n; p++) 
        { 
            if(factors[p] == 0) 
            { 
                factors[p]=p;
                for(int i = p*p; i <= n; i += p) 
                    factors[i] = p; 
            } 
        } 
    }

    static void sort(int ar[]) {
        int n = ar.length;
        ArrayList<Integer> a = new ArrayList<>();
        for (int i = 0; i < n; i++)
            a.add(ar[i]);
        Collections.sort(a);
        for (int i = 0; i < n; i++)
            ar[i] = a.get(i);
    }

    static void sort1(long ar[]) {
        int n = ar.length;
        ArrayList<Long> a = new ArrayList<>();
        for (int i = 0; i < n; i++)
            a.add(ar[i]);
        Collections.sort(a);
        for (int i = 0; i < n; i++)
            ar[i] = a.get(i);
    }

    static long ncr(long n, long r, long mod) {
        if (r == 0)
            return 1;
        long val = ncr(n - 1, r - 1, mod);
        val = (n * val) % mod;
        val = (val * modInverse(r, mod)) % mod;
        return val;
    }

    public static void solve(InputReader sc, PrintWriter pw) {
        int i, j = 0;
        int t = 1;  
        // int t = sc.nextInt();
        u: while (t-- > 0) {
            int n = sc.nextInt();
            int a[] = new int[n];
            for(i=0;i<n;i++){
                a[i] = sc.nextInt();
            }
            ArrayList<String> l = new ArrayList<>();
            for(i=0;i<n;i++){
                HashMap<Integer,Integer> map = new HashMap<>();
                int s = -1, g = -1;
                for(j=0;j<n;j++){
                    if(map.containsKey(a[j]+1)){
                        s = map.get(a[j]+1);
                        g = j;
                        // pw.println(a[j]+" "+s+" "+g);
                        break;
                    }
                    map.put(a[j],j);
                }
                // pw.println(s);
                if(s==-1)
                    break;
                int y = s;
                for(j=s+1;j<n;j++){
                    if(a[j]!=a[j-1]+1)
                        break;
                    y = j;
                }
                int z = g;
                // pw.println(s+" "+y+" "+z+" ");
                // for(j=g+1;j<n;j++){
                //     if(a[j]!=a[j-1]+1)
                //         break;
                //     z = j;
                // }
                int k = 0;
                String h = "";
                if(s>0){
                    k++;
                    h+=s+" ";
                }
                k++;
                h+=(y-s+1)+" ";
                k++;
                h+=(z-y)+" ";
                if(z<n-1){
                    k++;
                    h+=(n-1-z)+" ";
                }
                h = k+" "+h;
                l.add(h);
                int b[] = new int[n];
                int c = 0;
                for(j=g+1;j<n;j++){
                    b[c++] = a[j];
                }
                for(j=y+1;j<=g;j++){
                    b[c++] = a[j];
                }
                for(j=s;j<=y;j++){
                    b[c++] = a[j];
                }
                for(j=0;j<s;j++){
                    b[c++] = a[j];
                }
                for(j=0;j<n;j++){
                    a[j] = b[j];
                    // pw.println(j+" "+a[j]);
                }
            }
            pw.println(l.size());
            for(i=0;i<l.size();i++){
                pw.println(l.get(i));
            }
        }
    }

    static int bpm(ArrayList<Integer> adj[], int n){
        int pair[] = new int[2*n+1];
        int dist[] = new int[n+1];
        int match = 0;
        while(bfs(pair,dist,n,adj)){
            for(int v = 1; v <= n; v++){
                if(pair[v]==0 && dfs(v,pair,dist,n,adj)){
                    match++;
                }
            }
        }
        return match;
    }

    static boolean bfs(int pair[], int dist[], int n, ArrayList<Integer> adj[]){
        Deque<Integer> q = new LinkedList<>();
        for(int i=1;i<=n;i++){
            if(pair[i]==0){
                dist[i] = 0;
                q.offerLast(i);
            }
            else{
                dist[i] = Integer.MAX_VALUE;
            }
        }
        dist[0] = Integer.MAX_VALUE;
        while(q.size()>0){
            int v = q.pollFirst();
            if(dist[v]<dist[0]){
                for(int a:adj[v]){
                    if(dist[pair[a]]==Integer.MAX_VALUE){
                        dist[pair[a]] = dist[v] + 1;
                        q.offerLast(pair[a]);
                    }
                }
            }
        }
        return dist[0]!=Integer.MAX_VALUE;
    }

    static boolean dfs(int v, int pair[], int dist[], int n, ArrayList<Integer> adj[]){
        if(v==0)
            return true;
        for(int a:adj[v]){
            if(dist[pair[a]]==dist[v]+1){
                if(dfs(pair[a],pair,dist,n,adj)){
                    pair[a] = v;
                    pair[v] = a;
                    return true;
                }
            }
        }
        dist[v] = Integer.MAX_VALUE;
        return false;
    }

    static class Pair implements Comparable<Pair> {
        int a;
        int b;
        int c;
        
        Pair(int a, int b, int c) {
            this.a = a;
            this.b = b;
            this.c = c;
        }
 
        public int compareTo(Pair p) {
            // if(a!=p.a)
            //     return a-p.a;
            return c-p.c;
        }
    }

    static boolean isPrime(long n) {
        if (n <= 1)
            return false;
        if (n <= 3)
            return true;
        if (n % 2 == 0 || n % 3 == 0)
            return false;
        for (int i = 5; i * i <= n; i = i + 6)
            if (n % i == 0 || n % (i + 2) == 0)
                return false;
        return true;
    }

    static long gcd(long a, long b) {
        if (b == 0)
            return a;
        return gcd(b, a % b);
    }

    static long fast_pow(long base, long n, long M) {
        if (n == 0)
            return 1;
        if (n == 1)
            return base % M;
        long halfn = fast_pow(base, n / 2, M);
        if (n % 2 == 0)
            return (halfn * halfn) % M;
        else
            return (((halfn * halfn) % M) * base) % M;
    }

    static long modInverse(long n, long M) {
        return fast_pow(n, M - 2, M);
    }

    static class InputReader {
        public BufferedReader reader;
        public StringTokenizer tokenizer;

        public InputReader(InputStream stream) {
            reader = new BufferedReader(new InputStreamReader(stream), 32768);
            tokenizer = null;
        }

        public String next() {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }

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

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

