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 n, m, l, r, pos, val[11000], tmp_val[11000];
stack<int> ans[11000];
void sk(int l, int r) {
  if (r >= l && r - l + 1 < n) ans[m].push(r - l + 1);
  for (int j = l; j <= r; ++j) tmp_val[++pos] = val[j];
}
void work1(int i) {
  sk(n - i + 2, n);
  sk(r, n - i + 1);
  sk(l + 1, r - 1);
  sk(i, l);
  sk(1, i - 1);
}
void work2(int i) {
  sk(r + 1, n);
  if (val[l] < val[r])
    sk(l, r);
  else
    sk(r, r), sk(l + 1, r - 1), sk(l, l);
  sk(1, l - 1);
}
void work(int op, int i) {
  l = 1, r = n, pos = 0;
  while (val[l] != i && val[l] != n - i + 1) l++;
  while (val[r] != i && val[r] != n - i + 1) r--;
  m++;
  if (op == 1)
    work1(i);
  else
    work2(i);
  if (ans[m].size() == 0) m--;
  for (int j = 1; j <= n; ++j) val[j] = tmp_val[j];
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; ++i) cin >> val[i];
  for (int i = 1; 2 * i <= n; ++i) work(1, i), work(2, i);
  cout << m;
  for (int i = 1; i <= m; ++i) {
    cout << endl << ans[i].size() << " ";
    while (!ans[i].empty()) cout << ans[i].top() << " ", ans[i].pop();
  }
  return 0;
}