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 IO() {
  mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
int main() {
  IO();
  long long int n;
  cin >> n;
  vector<long long int> arr(n);
  for (__typeof(n) i = (0) - ((0) > (n)); i != (n) - ((0) > (n));
       i += 1 - 2 * ((0) > (n))) {
    cin >> arr[i];
  }
  vector<long long int> quantity(n, 1);
  long long int cp = 0;
  vector<vector<int> > ans;
  do {
    vector<long long int> brr;
    vector<long long int> nquant;
    n = (int)(arr.size());
    long long int prev = arr[0];
    set<int> put;
    put.insert(prev);
    long long int sum = quantity[0];
    for (__typeof(n) i = (1) - ((1) > (n)); i != (n) - ((1) > (n));
         i += 1 - 2 * ((1) > (n))) {
      long long int diff = arr[i] - prev;
      if (diff == 1) {
        put.insert(arr[i]);
        sum += quantity[i];
      } else {
        auto it = put.begin();
        brr.push_back(*it);
        nquant.push_back(sum);
        sum = quantity[i];
        put.clear();
        put.insert(arr[i]);
      }
      prev = arr[i];
    }
    auto it = put.begin();
    brr.push_back(*it);
    nquant.push_back(sum);
    vector<long long int> temp = brr;
    sort(temp.begin(), temp.end());
    map<long long int, long long int> seq;
    long long int cnt = 1;
    for (auto it : temp) {
      seq[it] = cnt;
      cnt++;
    }
    for (auto &it : brr) {
      it = seq[it];
    }
    arr = brr;
    quantity = nquant;
    if ((int)(arr.size()) > 1) {
      n = (int)(arr.size());
      long long int l = 0, r = 0;
      for (__typeof(n - 1) i = (0) - ((0) > (n - 1));
           i != (n - 1) - ((0) > (n - 1)); i += 1 - 2 * ((0) > (n - 1))) {
        for (__typeof(n) j = (i + 1) - ((i + 1) > (n));
             j != (n) - ((i + 1) > (n)); j += 1 - 2 * ((i + 1) > (n))) {
          long long int diff = arr[i] - arr[j];
          if (diff == 1) {
            l = i, r = j;
            break;
          }
        }
        if (r > l) break;
      }
      vector<int> res;
      vector<long long int> temp;
      vector<long long int> newQ;
      int cnt = 0;
      for (int i = r + 1; i <= n - 1; i++) {
        temp.push_back(arr[i]);
        newQ.push_back(quantity[i]);
        cnt += quantity[i];
      }
      if (cnt) res.push_back(cnt);
      temp.push_back(arr[r]);
      newQ.push_back(quantity[r]);
      res.push_back(quantity[r]);
      cnt = 0;
      for (int i = l; i <= r - 1; i++) {
        temp.push_back(arr[i]);
        newQ.push_back(quantity[i]);
        cnt += quantity[i];
      }
      if (cnt) res.push_back(cnt);
      cnt = 0;
      for (int i = 0; i <= l - 1; i++) {
        temp.push_back(arr[i]);
        newQ.push_back(quantity[i]);
        cnt += quantity[i];
      }
      if (cnt) res.push_back(cnt);
      reverse(res.begin(), res.end());
      arr = temp;
      quantity = newQ;
      ans.push_back(res);
    }
  } while ((int)(arr.size()) > 1);
  cout << (int)(ans.size()) << "\n";
  for (auto it : ans) {
    cout << (int)(it.size()) << " ";
    for (auto ti : it) cout << ti << " ";
    cout << "\n";
  }
  return 0;
}