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 main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n, a[52][52], b[52][52], ans[4][52], q = 0;
  bool yes = false;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i][0];
    b[i][0] = a[i][0];
    if (i != 0 && a[i - 1][0] > a[i][0]) yes = true;
  }
  while (yes) {
    int ind1, ind2, qan1 = 0, qan2 = 0, qan3 = 0;
    for (int k = 1; k < n; k++) {
      bool t = false;
      ind1 = -1, ind2 = -1;
      for (int i = 0; i < n - q; i++) {
        if (!t && b[i][q] == k) {
          ind2 = i;
          t = true;
        }
        if (!t && a[i][q] == k + 1) {
          ind1 = i;
        }
      }
      if (ind1 != -1 && ind2 != -1) {
        qan1 = 0, qan2 = 0, qan3 = 0;
        int sum = 0;
        for (int i = 0; i < ind1; i++) {
          sum += b[i][q] - a[i][q] + 1;
          qan1++;
        }
        ans[0][q] = sum;
        sum = 0;
        for (int i = ind1; i < ind2; i++) {
          sum += b[i][q] - a[i][q] + 1;
          qan2++;
        }
        ans[1][q] = sum;
        ans[2][q] = b[ind2][q] - a[ind2][q] + 1;
        sum = 0;
        for (int i = ind2 + 1; i < n - q; i++) {
          sum += b[i][q] - a[i][q] + 1;
          qan3++;
        }
        ans[3][q] = sum;
        k = n + 1;
      }
    }
    for (int i = 0; i < qan3; i++) {
      b[i][q + 1] = b[ind2 + i + 1][q];
      a[i][q + 1] = a[ind2 + i + 1][q];
    }
    a[qan3][q + 1] = a[ind2][q];
    b[qan3][q + 1] = b[ind1][q];
    for (int i = qan3 + 1; i < (qan3 + qan2); i++) {
      b[i][q + 1] = b[ind1 + (i - qan3 - 1) + 1][q];
      a[i][q + 1] = a[ind1 + (i - qan3 - 1) + 1][q];
    }
    for (int i = (qan3 + qan2); i < n - q - 1; i++) {
      b[i][q + 1] = b[i - qan3 - qan2][q];
      a[i][q + 1] = a[i - qan3 - qan2][q];
    }
    yes = false;
    for (int i = 0; i < n - q - 1; i++) {
      if (i != 0 && b[i - 1][q + 1] > a[i][q + 1]) yes = true;
    }
    q++;
  }
  cout << q << endl;
  for (int i = 0; i < q; i++) {
    int l = 0;
    for (int j = 0; j < 4; j++) {
      if (ans[j][i] != 0) l++;
    }
    cout << l << " ";
    for (int j = 0; j < 4; j++) {
      if (ans[j][i] != 0) cout << ans[j][i] << " ";
    }
    cout << endl;
  }
  return 0;
}