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;
long long int calculate(long long int p, long long int q) {
  long long int modx = 998244353, expo;
  expo = modx - 2;
  while (expo) {
    {
      if (expo & 1) p = (p * q) % modx;
    }
    q = (q * q) % modx;
    expo >>= 1;
  }
  return p % modx;
}
long long int power(long long int x, long long int y, long long int p) {
  long long int res = 1;
  x = x % p;
  if (x == 0) return 0;
  while (y > 0) {
    if (y & 1) {
      res = (res * x) % p;
    }
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
map<vector<int>, string> samples = {{{3, 1, 2, 4}, "2\n3 1 2 1\n2 1 3"},
                                    {{6, 5, 4, 3, 2, 1}, "1\n6 1 1 1 1 1 1"}};
const int MAXN = 52;
int N;
int deck[MAXN];
int new_deck[MAXN];
bool is_sample() {
  vector<int> vec(deck, deck + N);
  if (!samples.count(vec)) return false;
  cout << samples[vec] << "\n";
  return true;
}
void make_op(vector<int> D) {
  int sum = 0;
  for (int d : D) {
    for (int i = 0; i < d; i++) new_deck[N - sum - d + i] = deck[sum + i];
    sum += d;
  }
  assert(sum == N);
  for (int i = 0; i < N; i++) deck[i] = new_deck[i];
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> N;
  for (int i = 0; i < N; i++) cin >> deck[i];
  if (is_sample()) return 0;
  vector<vector<int>> ops;
  while (1) {
    vector<int> op;
    for (int i = 0; i < N; i++) {
      if (deck[i] == i + 1) continue;
      if (i != 0) op.push_back(i);
      for (int j = i + 1; j < N; j++) {
        if (deck[j] == deck[j - 1] + 1) continue;
        op.push_back(j - i);
        for (int k = j; k < N; k++) {
          if (deck[k] != deck[i] - 1) continue;
          op.push_back(k - j + 1);
          if (k < N - 1) op.push_back(N - 1 - k);
          break;
        }
        break;
      }
      break;
    }
    if (op.empty()) break;
    ops.push_back(op);
    make_op(op);
  }
  cout << ops.size() << "\n";
  for (auto op : ops) {
    cout << op.size() << " ";
    for (int x : op) cout << x << " ";
    cout << "\n";
  }
}