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;
const long long max_n = 55;
const long long inf = 1e18 + 10;
long long n, c[max_n];
vector<vector<long long> > operations;
void apply(vector<long long> D) {
  vector<vector<long long> > curr;
  long long i = 0;
  for (long long d : D) {
    vector<long long> t;
    for (long long w = 0; w < d; w++) {
      t.push_back(c[i++]);
    }
    curr.push_back(t);
  }
  reverse(curr.begin(), curr.end());
  i = 0;
  for (long long k = 0; k < curr.size(); k++) {
    for (long long x : curr[k]) {
      c[i] = x;
      i++;
    }
  }
  return;
}
bool do_forward() {
  long long i = 1;
  vector<long long> current;
  while (i < n && c[i] - c[i - 1] == 1) {
    current.push_back(1);
    i++;
  }
  current.push_back(1);
  i++;
  if (i > n) return 1;
  for (long long j = 0; j < n; j++) {
    if (c[j] == i) {
      current.push_back(j + 1 - (i - 1));
      long long rem = (n - (j + 1));
      if (rem > 0) current.push_back(rem);
      break;
    }
  }
  apply(current);
  operations.push_back(current);
  return 0;
}
bool do_backward() {
  long long i = n - 2;
  vector<long long> current;
  while (i >= 0 && c[i] - c[i + 1] == 1) {
    current.push_back(1);
    i--;
  }
  current.push_back(1);
  i = current.size() + 1;
  if (i > n) return 1;
  for (long long j = n - 1; j >= 0; j--) {
    if (c[j] == i) {
      current.push_back((n - j) - (i - 1));
      long long rem = (n - (n - j));
      if (rem > 0) current.push_back(rem);
      break;
    }
  }
  reverse(current.begin(), current.end());
  operations.push_back(current);
  apply(current);
  return 0;
}
void do_one() {
  vector<long long> current;
  for (long long j = 0; j < n; j++) {
    if (c[j] == 1) {
      if (j == 0) return;
      current.push_back(j);
      current.push_back(n - j);
    }
  }
  apply(current);
  operations.push_back(current);
  return;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n;
  for (long long i = 0; i < n; i++) cin >> c[i];
  do_one();
  bool toggle = 1;
  while (1) {
    if (toggle) {
      if (do_forward()) {
        break;
      }
    } else {
      if (do_backward()) {
        vector<long long> current;
        for (long long w = 0; w < n; w++) current.push_back(1);
        operations.push_back(current);
        apply(current);
        break;
      }
    }
    toggle = !toggle;
  }
  cout << operations.size() << "\n";
  for (vector<long long> o : operations) {
    cout << o.size() << " ";
    for (long long x : o) {
      cout << x << " ";
    }
    cout << "\n";
  }
}