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.*;
import java.io.*;
import java.math.*;
 
public class Main {
    static PrintWriter out;
    static Reader in;
    public static void main(String[] args) throws IOException {
        //out = new PrintWriter(new File("out.txt"));
        //PrintWriter out = new PrintWriter(System.out);
        //in = new Reader(new FileInputStream("in.txt"));
        //Reader in = new Reader();
        input_output();
        Main solver = new Main();
        solver.solve();
        out.flush();
        out.close();
 
    }
 
    static long INF = (long)1e16;
    static int maxn = (int)1e6+5;
    static int mod= 998_244_353;
    static int n, m, q, k, t;

    void solve() throws IOException{
        n = in.nextInt();

        int[] arr = new int[n];
        int[] pos = new int[n+1];
        for (int i = 0; i < n; i++) {
            arr[i] = in.nextInt();
            pos[arr[i]] = i;
        }

        if (n == 1) {
            out.println(0);
            return;
        }

        if (n == 2) {
            if (arr[0] == 1) out.println(0);
            else {
                out.println(1);
                out.println("2 1 1");
            }
            return;
        }

        boolean flag = true;
        for (int i = 1; i < n; i++) if (arr[i] < arr[i-1]) flag = false;
        if (flag ) {
            out.println(0);
            return;
        }

        int cnt = 0;
        ArrayList<Integer> ans = new ArrayList<>();

        while (true) {
            cnt++;

            int x = 0, y = 0;
            for (int i = 1; i < n; i++) {
                int j = i+1;
                if (pos[i] > pos[j]) {
                    x = pos[j];
                    y = pos[i];
                    break;
                }
            }

            int sizex = 1, sizey = 1;
            for (int i = x+1; i < n; i++) {
                if (arr[i]-1 == arr[i-1]) sizex++;
                else break;
            }

            ArrayList<Integer> flip = new ArrayList<>();
            sizey = y-x+1-sizex;

            int len = 2;
            if (x != 0) len++;
            if (y != n-1) len++;

            ans.add(len);
            if (x != 0) {
                ans.add(x);
                flip.add(x);
            }
            ans.add(sizex);
            flip.add(sizex);
            ans.add(sizey);
            flip.add(sizey);
            if (y != n-1) {
                ans.add(n-1-y);
                flip.add(n-1-y);
            }

            int[] tmp = new int[n];
            int idx = n-1, point = 0, cur = 0;
            for (int j = flip.size()-1; j >= 0; j--) {
                point = idx;
                idx = idx-flip.get(j)+1;
                for (int i = idx; i <= point; i++) {
                    tmp[cur++] = arr[i];
                }
                idx--;
            }
            for (int i = 0; i < n; i++) {
                arr[i] = tmp[i];
                pos[arr[i]] = i;
            }

            boolean sorted = true;
            for (int i = 1; i < n; i++) {
                if (arr[i] < arr[i-1]) sorted = false;
            }
            if (sorted) break;
        }

        out.println(cnt);
        int cur = 0;
        for (int i = 0; i < cnt; i++) {
            int len = ans.get(cur++);
            out.print(len);
            for (int j = cur; j < cur+len; j++) {
                out.print(" "+ans.get(j));
            }
            cur+=len;
            out.println();
        }

    }

    //<>

    static class Reader {
 
        private InputStream mIs;
        private byte[] buf = new byte[1024];
        private int curChar;
        private int numChars;
    
        public Reader() {
            this(System.in);
        }
    
        public Reader(InputStream is) {
            mIs = is;
        }
    
        public int read() {
            if (numChars == -1) {
                throw new InputMismatchException();
    
        }
            if (curChar >= numChars) {
                curChar = 0;
                try {
                    numChars = mIs.read(buf);
                } catch (IOException e) {
                    throw new InputMismatchException();
                }
                if (numChars <= 0) {
                    return -1;
                }
            }
            return buf[curChar++];
        }
    
        public String nextLine() {
            int c = read();
            while (isSpaceChar(c)) {
                c = read();
            }
            StringBuilder res = new StringBuilder();
            do {
                res.appendCodePoint(c);
                c = read();
            } while (!isEndOfLine(c));
            return res.toString();
        }
    
        public String next() {
            int c = read();
            while (isSpaceChar(c)) {
                c = read();
            }
            StringBuilder res = new StringBuilder();
            do {
                res.appendCodePoint(c);
                c = read();
            } while (!isSpaceChar(c));
            return res.toString();
        }
    
        double nextDouble()
        {
            return Double.parseDouble(next());
        }
    
        public long nextLong() {
            int c = read();
            while (isSpaceChar(c)) {
                c = read();
            }
            int sgn = 1;
            if (c == '-') {
                sgn = -1;
                c = read();
            }
            long res = 0;
            do {
                if (c < '0' || c > '9') {
                    throw new InputMismatchException();
                }
                res *= 10;
                res += c - '0';
                c = read();
            } while (!isSpaceChar(c));
            return res * sgn;
        }
    
        public int nextInt() {
            int c = read();
            while (isSpaceChar(c)) {
                c = read();
            }
            int sgn = 1;
            if (c == '-') {
                sgn = -1;
                c = read();
            }
            int res = 0;
            do {
                if (c < '0' || c > '9') {
                    throw new InputMismatchException();
                }
                res *= 10;
                res += c - '0';
                c = read();
            } while (!isSpaceChar(c));
            return res * sgn;
        }
    
        public boolean isSpaceChar(int c) {
            return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
        }
    
        public boolean isEndOfLine(int c) {
            return c == '\n' || c == '\r' || c == -1;
        }
 
    }
    static void input_output() throws IOException {
        File f = new File("in.txt");
        if(f.exists() && !f.isDirectory()) { 
            in = new Reader(new FileInputStream("in.txt"));
        } else in = new Reader();
        f = new File("out.txt");
        if(f.exists() && !f.isDirectory()) {
            out = new PrintWriter(new File("out.txt"));
        } else out = new PrintWriter(System.out);
    }
}