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;
void doop(vector<long long> &c, vector<long long> op) {
  long long n = c.size();
  vector<long long> tmp = c;
  long long j = 0;
  long long offset = n;
  for (auto opi : op) {
    offset -= opi;
    for (long long ji = j; ji < j + opi; ++ji) {
      tmp[offset + ji - j] = c[ji];
    }
    j += opi;
  }
  copy(tmp.begin(), tmp.end(), c.begin());
}
long long fi(vector<long long> &c, long long tf) {
  for (long long i = 0; i < (long long)c.size(); ++i) {
    if (c[i] == tf) {
      return i;
    }
  }
  return c.size();
}
void ij(vector<long long> &c, vector<vector<long long>> &ops, long long ss) {
  long long n = c.size();
  long long tf = (n / 2) + (n % 2 != 0);
  long long i = fi(c, tf);
  if (i == (long long)c.size() - 1) {
    return;
  }
  ops.push_back({i + 1, n - (i + 1)});
  doop(c, {i + 1, n - (i + 1)});
}
void lj(vector<long long> &c, vector<vector<long long>> &ops, long long ss) {
  long long n = c.size();
  long long tf = c[n - 1] + 1;
  long long i = fi(c, tf);
  if (i == 0) {
    ops.push_back({n - ss, ss});
    doop(c, {n - ss, ss});
  } else {
    ops.push_back({i, n - ss - i, ss});
    doop(c, {i, n - ss - i, ss});
  }
}
void rj(vector<long long> &c, vector<vector<long long>> &ops, long long ss) {
  long long n = c.size();
  long long tf = c[0] - 1;
  long long i = fi(c, tf);
  if (i == n - 1) {
    ops.push_back({ss, n - ss});
    doop(c, {ss, n - ss});
  } else {
    ops.push_back({ss, i + 1 - ss, n - (i + 1)});
    doop(c, {ss, i + 1 - ss, n - (i + 1)});
  }
}
int main(void) {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long n;
  cin >> n;
  vector<long long> c(n);
  for (auto &ci : c) {
    cin >> ci;
  }
  vector<vector<long long>> ops;
  long long ss = 0;
  ij(c, ops, ss);
  ss++;
  while (ss < n) {
    lj(c, ops, ss);
    ss++;
    if (ss >= n) {
      break;
    }
    rj(c, ops, ss);
    ss++;
  }
  cout << ops.size() << endl;
  for (auto opi : ops) {
    cout << opi.size() << " ";
    for (auto opij : opi) {
      cout << opij << " ";
    }
    cout << endl;
  }
  return 0;
}