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;
void solve() {
  int n;
  cin >> n;
  vector<int> A(n), B(n);
  for (int i = 0; i < n; i++) {
    cin >> A[i];
  }
  if (n == 1) {
    cout << "0\n";
    return;
  }
  int cnt = 0;
  vector<vector<int>> op(n);
  if (A[n - 1] != n) {
    for (int i = 0; i < n; i++) {
      if (A[i] == n) {
        op[cnt].push_back(i + 1);
        op[cnt].push_back(n - i - 1);
        for (int j = i + 1; j < n; j++) B[j - i - 1] = A[j];
        for (int j = 0; j <= i; j++) B[n - i - 1 + j] = A[j];
        swap(A, B);
        cnt++;
        break;
      }
    }
  }
  for (int i = n - 1; i >= 2; i--) {
    if (A[n - 1] == n) {
      if (A[0] == i) {
        op[cnt].push_back(i);
        for (int j = 0; j < n - i; j++) op[cnt].push_back(1);
        for (int j = n - 1; j >= i; j--) B[n - 1 - j] = A[j];
        for (int j = 0; j < i; j++) B[n - i + j] = A[j];
        swap(A, B);
        cnt++;
      } else {
        for (int j = 1; j < n; j++) {
          if (A[j] == i) {
            op[cnt].push_back(j);
            op[cnt].push_back(i - j);
            for (int k = 0; k < n - i; k++) op[cnt].push_back(1);
            for (int k = n - 1; k >= i; k--) B[n - 1 - k] = A[k];
            for (int k = j; k < i; k++) B[n - i + k - j] = A[k];
            for (int k = 0; k < j; k++) B[n - j + k] = A[k];
            swap(A, B);
            cnt++;
            break;
          }
        }
      }
    } else {
      if (A[n - 1] == i) {
        for (int j = 0; j < n - i; j++) op[cnt].push_back(1);
        op[cnt].push_back(i);
        for (int j = n - i; j < n; j++) B[j - n + i] = A[j];
        for (int j = n - i - 1; j >= 0; j--) B[n - 1 - j] = A[j];
        swap(A, B);
        cnt++;
      } else {
        for (int j = 0; j < n; j++) {
          if (A[j] == i) {
            for (int k = 0; k < n - i; k++) op[cnt].push_back(1);
            op[cnt].push_back(j + 1 - n + i);
            op[cnt].push_back(n - j - 1);
            for (int k = j + 1; k < n; k++) B[k - j - 1] = A[k];
            for (int k = n - i; k <= j; k++) B[-1 - j + k + i] = A[k];
            for (int k = n - i - 1; k >= 0; k--) B[n - 1 - k] = A[k];
            swap(A, B);
            cnt++;
            break;
          }
        }
      }
    }
  }
  if (A[0] == n) {
    int i = 1;
    if (A[n - 1] == i) {
      for (int j = 0; j < n - i; j++) op[cnt].push_back(1);
      op[cnt].push_back(i);
      for (int j = n - i; j < n; j++) B[j - n + i] = A[j];
      for (int j = n - i - 1; j >= 0; j--) B[n - 1 - j] = A[j];
      swap(A, B);
      cnt++;
    } else {
      for (int j = 0; j < n; j++) {
        if (A[j] == i) {
          for (int k = 0; k < n - i; k++) op[cnt].push_back(1);
          op[cnt].push_back(j + 1 - n + i);
          op[cnt].push_back(n - j - 1);
          for (int k = j + 1; k < n; k++) B[k - j - 1] = A[k];
          for (int k = n - i; k <= j; k++) B[-1 - j + k + i] = A[k];
          for (int k = n - i - 1; k >= 0; k--) B[n - 1 - k] = A[k];
          swap(A, B);
          cnt++;
          break;
        }
      }
    }
  }
  cout << cnt << '\n';
  for (int i = 0; i < cnt; i++) {
    cout << op[i].size();
    for (int x : op[i]) cout << ' ' << x;
    cout << '\n';
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie();
  int T = 1;
  while (T--) {
    solve();
  }
  return 0;
}