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.io.*;
import java.util.*;

import static java.lang.Math.*;
import static java.util.Arrays.*;

public class cf1427d {

    public static void main(String[] args) throws IOException {
        int n = ri(), a[] = riam1(n), ind[], mininv, mininvind;
        boolean found = true;
        List<List<Integer>> ans = new ArrayList<>();
        while (found) {
            found = false;
            ind = new int[n];
            fill(ind, -1);
            mininv = n;
            mininvind = -1;
            for (int i = 0; i < n; ++i) {
                ind[a[i]] = i;
                if (a[i] > 0 && ind[a[i] - 1] == -1 && a[i] < mininv) {
                    mininvind = i;
                    mininv = a[i];
                    found = true;
                }
            }
            // prln(mininv, mininvind);
            if (found) {
                int na[] = new int[n], ii = 0, cut = ind[mininv - 1];
                while (a[cut - 1] == a[cut] - 1) {
                    --cut;
                }
                List<Integer> e = new ArrayList<>();
                if (mininvind > 0) {
                    e.add(mininvind);
                }
                e.add(cut - mininvind);
                e.add(ind[mininv - 1] - cut + 1);
                if (ind[mininv - 1] < n - 1) {
                    e.add(n - ind[mininv - 1] - 1);
                }
                for (int i = ind[mininv - 1] + 1; i < n; ++i) {
                    na[ii++] = a[i];
                }
                for (int i = cut; i <= ind[mininv - 1]; ++i) {
                    na[ii++] = a[i];
                }
                for (int i = mininvind; i < cut; ++i) {
                    na[ii++] = a[i];
                }
                for (int i = 0; i < mininvind; ++i) {
                    na[ii++] = a[i];
                }
                ans.add(e);
                a = na;
            }
        }
        prln(ans.size());
        for (List<Integer> e : ans) {
            pr(e.size());
            pr(' ');
            prln(e);
        }
        close();
    }

    static BufferedReader __in = new BufferedReader(new InputStreamReader(System.in));
    static PrintWriter __out = new PrintWriter(new OutputStreamWriter(System.out));
    static StringTokenizer input;
    static Random __rand = new Random();

    // references
    // IBIG = 1e9 + 7
    // IMAX ~= 2e9
    // LMAX ~= 9e18

    // constants
    static final int IBIG = 1000000007;
    static final int IMAX = 2147483647;
    static final int IMIN = -2147483648;
    static final long LMAX = 9223372036854775807L;
    static final long LMIN = -9223372036854775808L;
    // math util
    static int minof(int a, int b, int c) {return min(a, min(b, c));}
    static int minof(int... x) {if (x.length == 1) return x[0]; if (x.length == 2) return min(x[0], x[1]); if (x.length == 3) return min(x[0], min(x[1], x[2])); int min = x[0]; for (int i = 1; i < x.length; ++i) if (x[i] < min) min = x[i]; return min;}
    static long minof(long a, long b, long c) {return min(a, min(b, c));}
    static long minof(long... x) {if (x.length == 1) return x[0]; if (x.length == 2) return min(x[0], x[1]); if (x.length == 3) return min(x[0], min(x[1], x[2])); long min = x[0]; for (int i = 1; i < x.length; ++i) if (x[i] < min) min = x[i]; return min;}
    static int maxof(int a, int b, int c) {return max(a, max(b, c));}
    static int maxof(int... x) {if (x.length == 1) return x[0]; if (x.length == 2) return max(x[0], x[1]); if (x.length == 3) return max(x[0], max(x[1], x[2])); int max = x[0]; for (int i = 1; i < x.length; ++i) if (x[i] > max) max = x[i]; return max;}
    static long maxof(long a, long b, long c) {return max(a, max(b, c));}
    static long maxof(long... x) {if (x.length == 1) return x[0]; if (x.length == 2) return max(x[0], x[1]); if (x.length == 3) return max(x[0], max(x[1], x[2])); long max = x[0]; for (int i = 1; i < x.length; ++i) if (x[i] > max) max = x[i]; return max;}
    static int powi(int a, int b) {if (a == 0) return 0; int ans = 1; while (b > 0) {if ((b & 1) > 0) ans *= a; a *= a; b >>= 1;} return ans;}
    static long powl(long a, int b) {if (a == 0) return 0; long ans = 1; while (b > 0) {if ((b & 1) > 0) ans *= a; a *= a; b >>= 1;} return ans;}
    static int fli(double d) {return (int) d;}
    static int cei(double d) {return (int) ceil(d);}
    static long fll(double d) {return (long) d;}
    static long cel(double d) {return (long) ceil(d);}
    static int gcf(int a, int b) {return b == 0 ? a : gcf(b, a % b);}
    static long gcf(long a, long b) {return b == 0 ? a : gcf(b, a % b);}
    static int lcm(int a, int b) {return a * b / gcf(a, b);}
    static long lcm(long a, long b) {return a * b / gcf(a, b);}
    static int randInt(int min, int max) {return __rand.nextInt(max - min + 1) + min;}
    static long mix(long x) {x += 0x9e3779b97f4a7c15L; x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9L; x = (x ^ (x >> 27)) * 0x94d049bb133111ebL; return x ^ (x >> 31);}
    // array util
    static void reverse(int[] a) {for (int i = 0, n = a.length, half = n / 2; i < half; ++i) {int swap = a[i]; a[i] = a[n - i - 1]; a[n - i - 1] = swap;}}
    static void reverse(long[] a) {for (int i = 0, n = a.length, half = n / 2; i < half; ++i) {long swap = a[i]; a[i] = a[n - i - 1]; a[n - i - 1] = swap;}}
    static void reverse(double[] a) {for (int i = 0, n = a.length, half = n / 2; i < half; ++i) {double swap = a[i]; a[i] = a[n - i - 1]; a[n - i - 1] = swap;}}
    static void reverse(char[] a) {for (int i = 0, n = a.length, half = n / 2; i < half; ++i) {char swap = a[i]; a[i] = a[n - i - 1]; a[n - i - 1] = swap;}}
    static void shuffle(int[] a) {int n = a.length - 1; for (int i = 0; i < n; ++i) {int ind = randInt(i, n); int swap = a[i]; a[i] = a[ind]; a[ind] = swap;}}
    static void shuffle(long[] a) {int n = a.length - 1; for (int i = 0; i < n; ++i) {int ind = randInt(i, n); long swap = a[i]; a[i] = a[ind]; a[ind] = swap;}}
    static void shuffle(double[] a) {int n = a.length - 1; for (int i = 0; i < n; ++i) {int ind = randInt(i, n); double swap = a[i]; a[i] = a[ind]; a[ind] = swap;}}
    static void rsort(int[] a) {shuffle(a); sort(a);}
    static void rsort(long[] a) {shuffle(a); sort(a);}
    static void rsort(double[] a) {shuffle(a); sort(a);}
    static int[] copy(int[] a) {int[] ans = new int[a.length]; for (int i = 0; i < a.length; ++i) ans[i] = a[i]; return ans;}
    static long[] copy(long[] a) {long[] ans = new long[a.length]; for (int i = 0; i < a.length; ++i) ans[i] = a[i]; return ans;}
    static double[] copy(double[] a) {double[] ans = new double[a.length]; for (int i = 0; i < a.length; ++i) ans[i] = a[i]; return ans;}
    static char[] copy(char[] a) {char[] ans = new char[a.length]; for (int i = 0; i < a.length; ++i) ans[i] = a[i]; return ans;}
    // input
    static void r() throws IOException {input = new StringTokenizer(rline());}
    static int ri() throws IOException {return Integer.parseInt(rline());}
    static long rl() throws IOException {return Long.parseLong(rline());}
    static double rd() throws IOException {return Double.parseDouble(rline());}
    static int[] ria(int n) throws IOException {int[] a = new int[n]; r(); for (int i = 0; i < n; ++i) a[i] = ni(); return a;}
    static int[] riam1(int n) throws IOException {int[] a = new int[n]; r(); for (int i = 0; i < n; ++i) a[i] = ni() - 1; return a;}
    static long[] rla(int n) throws IOException {long[] a = new long[n]; r(); for (int i = 0; i < n; ++i) a[i] = nl(); return a;}
    static double[] rda(int n) throws IOException {double[] a = new double[n]; r(); for (int i = 0; i < n; ++i) a[i] = nd(); return a;}
    static char[] rcha() throws IOException {return rline().toCharArray();}
    static String rline() throws IOException {return __in.readLine();}
    static String n() {return input.nextToken();}
    static int rni() throws IOException {r(); return ni();}
    static int ni() {return Integer.parseInt(n());}
    static long rnl() throws IOException {r(); return nl();}
    static long nl() {return Long.parseLong(n());}
    static double rnd() throws IOException {r(); return nd();}
    static double nd() {return Double.parseDouble(n());}
    // output
    static void pr(int i) {__out.print(i);}
    static void prln(int i) {__out.println(i);}
    static void pr(long l) {__out.print(l);}
    static void prln(long l) {__out.println(l);}
    static void pr(double d) {__out.print(d);}
    static void prln(double d) {__out.println(d);}
    static void pr(char c) {__out.print(c);}
    static void prln(char c) {__out.println(c);}
    static void pr(char[] s) {__out.print(new String(s));}
    static void prln(char[] s) {__out.println(new String(s));}
    static void pr(String s) {__out.print(s);}
    static void prln(String s) {__out.println(s);}
    static void pr(Object o) {__out.print(o);}
    static void prln(Object o) {__out.println(o);}
    static void prln() {__out.println();}
    static void pryes() {prln("yes");}
    static void pry() {prln("Yes");}
    static void prY() {prln("YES");}
    static void prno() {prln("no");}
    static void prn() {prln("No");}
    static void prN() {prln("NO");}
    static boolean pryesno(boolean b) {prln(b ? "yes" : "no"); return b;};
    static boolean pryn(boolean b) {prln(b ? "Yes" : "No"); return b;}
    static boolean prYN(boolean b) {prln(b ? "YES" : "NO"); return b;}
    static void prln(int... a) {for (int i = 0, len = a.length - 1; i < len; pr(a[i]), pr(' '), ++i); if (a.length > 0) prln(a[a.length - 1]); else prln();}
    static void prln(long... a) {for (int i = 0, len = a.length - 1; i < len; pr(a[i]), pr(' '), ++i); if (a.length > 0) prln(a[a.length - 1]); else prln();}
    static void prln(double... a) {for (int i = 0, len = a.length - 1; i < len; pr(a[i]), pr(' '), ++i); if (a.length > 0) prln(a[a.length - 1]); else prln();}
    static <T> void prln(Collection<T> c) {int n = c.size() - 1; Iterator<T> iter = c.iterator(); for (int i = 0; i < n; pr(iter.next()), pr(' '), ++i); if (n >= 0) prln(iter.next()); else prln();}
    static void h() {prln("hlfd"); flush();}
    static void flush() {__out.flush();}
    static void close() {__out.close();}
}