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 x[100];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> x[i];
  if (n == 1) {
    cout << 0 << endl;
    return 0;
  } else if (n == 2) {
    if (x[1] == 1 && x[2] == 2) {
      cout << 0 << endl;
    } else
      cout << "1\n2 1 1\n";
    return 0;
  }
  vector<vector<int> > vec;
  int l = (n + 1) / 2, r = l + 1, flg = 0, ans_flg = 0;
  for (int i = 1; i <= n; i++) {
    int now;
    if (!flg)
      now = l--;
    else
      now = r++;
    flg ^= 1;
    for (int j = 1; j <= n; j++) {
      if (x[j] == now) {
        int a = i - 1;
        int b = j - i + 1;
        int c = n - j;
        vector<int> tmp;
        if (a != 0) tmp.push_back(a);
        if (b != 0) tmp.push_back(b);
        if (c != 0) tmp.push_back(c);
        if (tmp.size() == 1) {
          tmp.clear();
          for (int i = 1; i <= n; i++) tmp.push_back(1);
          ans_flg = 1;
          reverse(x + 1, x + n + 1);
        } else {
          reverse(x + 1, x + a + 1);
          reverse(x + a + 1, x + a + b + 1);
          reverse(x + a + b + 1, x + a + b + c + 1);
          vec.push_back(tmp);
        }
      }
    }
  }
  cout << vec.size() << endl;
  for (auto &it : vec) {
    cout << it.size() << " ";
    if (ans_flg) reverse(it.begin(), it.end());
    for (auto &it2 : it) cout << it2 << " ";
    cout << endl;
    ans_flg ^= 1;
  }
}