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 int M = 2e5 + 10, inf = 2e9 + 10, mod = 1e9 + 7, inv2 = (mod + 1) / 2;
int n, c[55], pos[55], c1[55];
void f(int l, int r, int num) {
  for (int i = l; i <= r; ++i) {
    int t = c[i];
    c1[i + num] = t;
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    cin >> c[i];
    c1[i] = c[i];
    pos[c[i]] = i;
  }
  int flag = n & 1;
  vector<vector<int>> res;
  for (int i = 1; i <= n; ++i) {
    int id = pos[i];
    vector<int> ans;
    if (flag == 0) {
      for (int j = i - 1; j > 0; --j) {
        ans.push_back(1);
        c1[n - j + 1] = j;
      }
      f(i, id, n - i - id + 1);
      f(id + 1, n, -id);
      if (id - i + 1) ans.push_back(id - i + 1);
      if (n - id) ans.push_back(n - id);
      if (ans.size() > 1) {
        res.push_back(ans);
      }
    } else {
      f(id, n - i + 1, i - id);
      f(1, id - 1, n - id + 1);
      if (id - 1) ans.push_back(id - 1);
      if (n - i - id + 2) ans.push_back(n - i - id + 2);
      for (int j = i - 1; j > 0; --j) {
        ans.push_back(1);
        c1[j] = j;
      }
      if (ans.size() > 1) {
        res.push_back(ans);
      }
    }
    for (int j = 1; j <= n; ++j) {
      c[j] = c1[j];
      pos[c[j]] = j;
    }
    flag ^= 1;
  }
  cout << res.size() << '\n';
  for (auto x : res) {
    cout << x.size() << ' ';
    for (int i = 0; i < x.size(); ++i) cout << x[i] << " \n"[i == x.size() - 1];
  }
}