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;
int const INF = (int)1e9 + 1e3;
long long const INFL = (long long)1e18 + 1e6;
mt19937 tw(chrono::high_resolution_clock::now().time_since_epoch().count());
uniform_int_distribution<long long> ll_distr;
long long rnd(long long a, long long b) {
  return ll_distr(tw) % (b - a + 1) + a;
}
void solve() {
  int n;
  cin >> n;
  vector<int> a(n), b(n);
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    a[i]--;
  }
  vector<vector<int>> ans;
  for (int i = 0; i < n; i++) {
    vector<int> cur;
    if (a[0] == 0) {
      int k = 0;
      while (k < n && a[k] == k) k++;
      if (k == n) break;
      int l = n - 1;
      while (l >= 0 && a[l] == l) l--;
      int q;
      for (int j = 0; j < n; j++)
        if (a[j] == l) q = j;
      for (int j = 0; j < k; j++) cur.push_back(1);
      if (q != k) {
        cur.push_back(q - k);
      }
      cur.push_back(l + 1 - q);
      for (int j = 0; j < n - 1 - l; j++) cur.push_back(1);
      int last = 0;
      for (int j = n - 1; j > l; j--) b[last++] = a[j];
      for (int j = q; j <= l; j++) b[last++] = a[j];
      for (int j = k; j < q; j++) b[last++] = a[j];
      for (int j = k - 1; j >= 0; j--) b[last++] = a[j];
    } else {
      int k = 0;
      while (k < n && a[k] == n - 1 - k) k++;
      if (k == n) {
        reverse(a.begin(), a.end());
        for (int j = 0; j < n; j++) cur.push_back(1);
        ans.push_back(cur);
        break;
      }
      int l = n - 1;
      while (l >= 0 && a[l] == n - 1 - l) l--;
      int q;
      for (int j = 0; j < n; j++)
        if (a[j] == n - 1 - l) q = j;
      for (int j = 0; j < k; j++) cur.push_back(1);
      if (q != k) {
        cur.push_back(q - k);
      }
      cur.push_back(l + 1 - q);
      for (int j = 0; j < n - 1 - l; j++) cur.push_back(1);
      int last = 0;
      for (int j = n - 1; j > l; j--) b[last++] = a[j];
      for (int j = q; j <= l; j++) b[last++] = a[j];
      for (int j = k; j < q; j++) b[last++] = a[j];
      for (int j = k - 1; j >= 0; j--) b[last++] = a[j];
    }
    for (int j = 0; j < n; j++) {
      a[j] = b[j];
    }
    ans.push_back(cur);
  }
  cout << ans.size() << "\n";
  for (int i = 0; i < (int)ans.size(); i++) {
    cout << ans[i].size() << " ";
    for (int j : ans[i]) cout << j << " ";
    cout << "\n";
  }
}
int main() {
  cout << setprecision(15) << fixed;
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int test_count = 1;
  for (int test = 1; test <= test_count; ++test) {
    solve();
  }
}