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>
#pragma comment(linker, "/STACK:512000000")
using namespace std;
void solve(__attribute__((unused)) bool);
void precalc();
clock_t start;
int main() {
  start = clock();
  int t = 1;
  cout.sync_with_stdio(0);
  cin.tie(0);
  precalc();
  cout.precision(10);
  cout << fixed;
  int testNum = 1;
  while (t--) {
    solve(true);
  }
  cout.flush();
  return 0;
}
template <typename T>
T binpow(T q, T w, T mod) {
  if (!w) return 1 % mod;
  if (w & 1) return q * 1LL * binpow(q, w - 1, mod) % mod;
  return binpow(q * 1LL * q % mod, w / 2, mod);
}
template <typename T>
T gcd(T q, T w) {
  while (w) {
    q %= w;
    swap(q, w);
  }
  return q;
}
template <typename T>
T lcm(T q, T w) {
  return q / gcd(q, w) * w;
}
template <typename T>
void make_unique(vector<T>& a) {
  sort(a.begin(), a.end());
  a.erase(unique(a.begin(), a.end()), a.end());
}
template <typename T>
void relax_min(T& cur, T val) {
  cur = min(cur, val);
}
template <typename T>
void relax_max(T& cur, T val) {
  cur = max(cur, val);
}
void precalc() {}
void solve(__attribute__((unused)) bool read) {
  int n;
  cin >> n;
  vector<int> p(n);
  for (int& x : p) {
    cin >> x;
    --x;
  }
  vector<vector<int>> ans;
  auto perform = [&](vector<int> d) {
    while (!d.empty() && d.back() == 0) {
      d.pop_back();
    }
    assert(!d.empty());
    if (d.size() == 1) {
      assert(d[0] == n);
      return;
    }
    ans.push_back(d);
    reverse(d.begin(), d.end());
    vector<int> res_p;
    for (int x : d) {
      vector<int> vec;
      for (int j = 0; j < x; ++j) {
        vec.push_back(p.back());
        p.pop_back();
      }
      reverse(vec.begin(), vec.end());
      for (int x : vec) {
        res_p.push_back(x);
      }
    }
    p.swap(res_p);
  };
  for (int have = 0; have + 1 < n; ++have) {
    if (have % 2 == 0) {
      for (int i = 0; i < n; ++i) {
        if (p[i] == have) {
          vector<int> ops(have, 1);
          ops.push_back(i + 1 - have);
          ops.push_back(n - i - 1);
          perform(ops);
          break;
        }
      }
    } else {
      for (int i = 0; i < n; ++i) {
        if (p[i] == have) {
          vector<int> ops;
          if (i) {
            ops.push_back(i);
          }
          ops.push_back(n - have - i);
          for (int j = 0; j < have; ++j) {
            ops.push_back(1);
          }
          perform(ops);
          break;
        }
      }
    }
  }
  if (p[0] != 0) {
    perform(vector<int>(n, 1));
  }
  for (int i = 0; i < n; ++i) {
    assert(p[i] == i);
  }
  assert(ans.size() <= n);
  cout << ans.size() << "\n";
  for (auto& vec : ans) {
    cout << vec.size() << " ";
    for (int x : vec) {
      cout << x << " ";
    }
    cout << "\n";
  }
}