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 CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
using namespace std::chrono;
long long fexp(long long a, long long b, long long m) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = ans * a % m;
    b /= 2;
    a = a * a % m;
  }
  return ans;
}
long long inverse(long long a, long long p) { return fexp(a, p - 2, p); }
long long fac[300005];
long long ncr(long long n, long long r, long long p) {
  if (r == 0) return 1;
  return (fac[n] * inverse(fac[r], p) % p * inverse(fac[n - r], p) % p) % p;
}
void fun(vector<long long> &v, vector<long long> &u) {
  long long a = (long long)u.size();
  vector<vector<long long>> w(a);
  long long k = 0;
  for (long long i = 0; i < a; i++) {
    for (long long j = 0; j < u[i]; j++) {
      w[i].push_back(v[k]);
      k++;
    }
  }
  k = 0;
  for (long long i = a - 1; i >= 0; i--) {
    for (long long j = 0; j < u[i]; j++) {
      v[k] = w[i][j];
      k++;
    }
  }
}
long long solve() {
  long long n;
  cin >> n;
  vector<long long> a(n);
  for (long long i = 0; i < n; i++) cin >> a[i];
  vector<vector<long long>> v;
  long long f = 0;
  for (long long i = 0; i < n - 1; i++) {
    vector<long long> u;
    long long x = 0, y = 0;
    for (long long j = 0; j < n; j++) {
      if (a[j] == 1) x = j;
    }
    for (long long j = 0; j < n; j++) {
      if (a[j] == i + 2) y = j;
    }
    if (f == 0 && x > y) {
      if (y > 0) u.push_back(y);
      u.push_back(x - y);
      u.push_back(i + 1);
      long long z = n - (x + i + 1);
      if (z > 0) u.push_back(z);
    } else if (f == 0 && x < y) {
      if (x > 0) u.push_back(x);
      for (long long j = 0; j < i + 1; j++) u.push_back(1);
      u.push_back(y - x - i);
      if (n - y - 1 > 0) u.push_back(n - 1 - y);
      f = 1;
    } else if (f == 1 && x > y) {
      if (y > 0) u.push_back(y);
      u.push_back(x - y - i);
      for (long long j = 0; j < i + 1; j++) u.push_back(1);
      if (n - x - 1 > 0) u.push_back(n - x - 1);
      f = 0;
    } else if (f == 1 && x < y) {
      if (x - i > 0) u.push_back(x - i);
      u.push_back(i + 1);
      u.push_back(y - x);
      if (n - y - 1 > 0) u.push_back(n - 1 - y);
    }
    fun(a, u);
    v.push_back(u);
  }
  if (f == 1) {
    vector<long long> u;
    for (long long i = 0; i < n; i++) u.push_back(1);
    v.push_back(u);
  }
  cout << (long long)v.size() << "\n";
  for (auto vv : v) {
    cout << (long long)vv.size() << " ";
    for (auto vvv : vv) cout << vvv << " ";
    cout << "\n";
  }
  return 0;
}
signed main() {
  auto start = high_resolution_clock::now();
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  long long t = 1;
  while (t--) {
    solve();
    cout << "\n";
  }
  auto stop = high_resolution_clock::now();
  auto duration = duration_cast<microseconds>(stop - start);
  cerr << "Time taken:" << duration.count() / 1000000.0 << "seconds"
       << "\n";
  return 0;
}