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.lang.*;
// StringBuilder uses java.lang
 
 
public class mC {
    
    
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        StringBuilder st = new StringBuilder();
        int t = 1;
        for (int test = 0; test < t; test++) {
            int n = sc.nextInt();
            int[] u = new int[n];
            int[] at = new int[n];
            for (int i=0;i<n;i++) {
                u[i]=sc.nextInt()-1;
                at[u[i]]=i;
            }
            int endGood = 0;
            for (int i=n-1;i>=0;i--) {
                if (i==u[i]) {
                    endGood++;
                } else {
                    break;
                }
            }
            int beginGood = 0;
            for (int i=0;i<n;i++) {
                if (n-i-1==u[i]) {
                    beginGood++;
                } else {
                    break;
                }
            }
            int numOps = 0;
            boolean sortedPM = false;
            if (endGood==n || beginGood==n) {
                sortedPM = true;
            }
            int k = 0;
            while (sortedPM==false) {
                numOps++;
                int r = goodRight(n,u);
                int l = goodLeft(n,u);
                if (r>0) { // list is 0-indexed
                    int a=n-r-1;
                    int o = at[a];
                    // o,n-o-r,1*r
                    if (o>0) {
                        st.append((r+2)+" "+o+" "+(n-o-r));
                    } else {
                        st.append((r+1)+" "+(n-r));
                    } 
                    for (int ind=0;ind<r;ind++) {
                        st.append(" 1");
                    }
                    st.append("\n");
                    /// doing operation
                    int[] v = new int[n];
                    for (int i=0;i<r;i++) {
                        v[i]=u[n-i-1];
                    }
                    for (int i=0;i<(n-o-r);i++) {
                        v[i+r]=u[o+i];
                    }
                    for (int i=0;i<o;i++) {
                        v[i+n-o]=u[i];
                    }
                    ////
                    for (int i=0;i<n;i++) {
                        u[i]=v[i];
                        at[u[i]]=i;
 
                    }
                    k+=1;
                    
                } else {
                    int b = n-l-1;
                    int p = at[b];
                    if (n-p-1==0) {
                        st.append((l+1)+" ");
                    } else {
                        st.append((l+2)+" ");
                    }
                    for (int ind=0;ind<l;ind++) {
                        st.append("1 ");
                    }
                    st.append((p-l+1));
                    if (n-p-1!=0) {
                        st.append(" "+(n-p-1));
                    }
                    st.append("\n");
 
                    ///
                    int[] v = new int[n];
                    for (int i=0;i<(n-p-1);i++) {
                        v[i] = u[n-(n-p-1)+i];
                    }
                    for (int i=0;i<(p-l+1);i++) {
                        v[i+(n-p-1)]=u[i+l];
                    }
                    for (int i=0;i<l;i++) {
                        v[n-i-1]=u[i];
                    }
                    ///
                    for (int i=0;i<n;i++) {
                        u[i]=v[i];
 
                        at[u[i]]=i;
                    }
                    k+=1;
 
                }
                
                if (goodRight(n,u)==n || goodLeft(n,u)==n) {
                    sortedPM=true;
                }
            }
            
            if (goodRight(n,u)==0) {
                numOps++;
                st.append(n);
                for (int i=0;i<n;i++) {
                    st.append(" 1");
                }
                st.append("\n");
            }
            System.out.println(numOps);
            System.out.print(st.toString());
        }
            
        
        
    }
    public static int goodLeft(int n, int[] p) { // i.e. ... 4 5 6
        int begin = 0;
        for (int i=0;i<n;i++) {
            if (n-i-1==p[i]) {
                begin++;
            } else {
                break;
            }
        }
        return begin;
    }
    public static int goodRight(int n, int[] p) { // i.e. 6 5 4 ...
        int end = 0;
        for (int i=n-1;i>=0;i--) {
            if (i==p[i]) {
                end++;
            } else {
                break;
            }
        }
        return end;
    }
    public static int findNthInArray(int[] arr,int val,int start,int o) {
        if (o==0) {
            return start-1;
        } else if (arr[start] == val) {
            return findNthInArray(arr,val,start+1,o-1);
        } else {
            return findNthInArray(arr,val,start+1,o);
        }
    }
    public static ArrayList<Integer> dfs(int at,ArrayList<Integer> went,ArrayList<ArrayList<Integer>> connect) {
        for (int i=0;i<connect.get(at).size();i++) {
            if (!(went.contains(connect.get(at).get(i)))) {
                went.add(connect.get(at).get(i));
                went=dfs(connect.get(at).get(i),went,connect);
            }
        }
        return went;
    } public static int[] bfs (int at, int[] went, ArrayList<ArrayList<Integer>> queue, int numNodes, ArrayList<ArrayList<Integer>> connect) {
        if (went[at]==0) {
            went[at]=queue.get(numNodes).get(1);
            for (int i=0;i<connect.get(at).size();i++) {
                if (went[connect.get(at).get(i)]==0) {
                    ArrayList<Integer> temp = new ArrayList<>();
                    temp.add(connect.get(at).get(i));
                    temp.add(queue.get(numNodes).get(1)+1);
                    queue.add(temp);
                }
            }
            
        }
        if (queue.size()==numNodes+1) {
            return went;
        } else {
            return bfs(queue.get(numNodes+1).get(0),went, queue, numNodes+1, connect);
        }
    }
    public static long fastPow(long base,long exp,long mod) {
        if (exp==0) {
            return 1;
        } else {
            if (exp % 2 == 1) {
                long z = fastPow(base,(exp-1)/2,mod);
                return ((((z*base) % mod) * z) % mod);
            } else {
                long z = fastPow(base,exp/2,mod);
                return ((z*z) % mod);
            }
        }
    }
    public static long fastPow(long base,long exp) {
        if (exp==0) {
            return 1;
        } else {
            if (exp % 2 == 1) {
                long z = fastPow(base,(exp-1)/2);
                return ((((z*base)) * z));
            } else {
                long z = fastPow(base,exp/2);
                return ((z*z));
            }
        }
    }
    public static int firstLarger(int val,ArrayList<Integer> ok,int left,int right) {
        if (ok.get(right)<=val) {
            return -1;
        }
        if (left==right) {
            return left;
        } else if (left+1==right) {
            if (val<ok.get(left)) {
                return left;
            } else {
                return right;
            }
        } else {
            int mid = (left+right)/2;
            if (ok.get(mid)>val) {
                return firstLarger(val,ok,left,mid);
            } else {
                return firstLarger(val,ok,mid+1,right);
            }
        }
    }
}