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;
vector<int> a, Q;
vector<vector<int>> Q2;
int n;
void help(int x) {
  int p1, p2, p3, s1, s2, s3, s4;
  vector<int> A, B, C, D;
  p3 = find(a.begin(), a.end(), x) - a.begin();
  p1 = find(a.begin(), a.end(), x + 1) - a.begin() - 1;
  p2 = p3;
  while (a[p2] != x + 1) --p2;
  while (a[p2 + 1] == a[p2] + 1) ++p2;
  for (int i = 1; i <= p1; ++i) A.push_back(a[i]);
  for (int i = p1 + 1; i <= p2; ++i) B.push_back(a[i]);
  for (int i = p2 + 1; i <= p3; ++i) C.push_back(a[i]);
  for (int i = p3 + 1; i <= n; ++i) D.push_back(a[i]);
  int sz = (!A.empty()) + (!B.empty()) + (!C.empty()) + (!D.empty());
  if (sz) {
    Q.push_back(sz);
    vector<int> temp;
    if (!A.empty()) temp.push_back(A.size());
    if (!B.empty()) temp.push_back(B.size());
    if (!C.empty()) temp.push_back(C.size());
    if (!D.empty()) temp.push_back(D.size());
    Q2.push_back(temp);
  }
  vector<int> b{0};
  for (auto it : D) b.push_back(it);
  for (auto it : C) b.push_back(it);
  for (auto it : B) b.push_back(it);
  for (auto it : A) b.push_back(it);
  a = b;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  a = vector<int>(n + 1);
  for (int i = 1; i <= n; ++i) cin >> a[i];
  while (1) {
    int no = 0;
    for (int j = 1; j < n; ++j) {
      if (find(a.begin(), a.end(), j + 1) < find(a.begin(), a.end(), j)) {
        no = j;
        break;
      }
    }
    if (!no) break;
    help(no);
  }
  cout << Q.size() << '\n';
  for (int i = 0; i < Q.size(); ++i, cout << '\n') {
    cout << Q[i] << ' ';
    for (auto it : Q2[i]) cout << it << ' ';
  }
}