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, num[100], temp[100], r[150], op[150][150], ans;
bool check() {
  for (int i = 1; i <= n; i++) {
    if (num[i] != i) return false;
  }
  return true;
}
bool check1() {
  for (int i = 1; i < n; i++) {
    if (num[i + 1] != num[i] - 1) return false;
  }
  return true;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &num[i]);
  }
  if (check()) {
    printf("0");
    return 0;
  }
  while (!check()) {
    ans++;
    if (check1()) {
      r[ans] = n;
      for (int i = 0; i < n; i++) op[ans][i] = 1;
      break;
    }
    int p1, p2;
    bool flag = true;
    for (int i = 1; i < n && flag; i++) {
      for (int j = 1; j <= n && flag; j++) {
        for (int k = 1; k <= n && flag; k++) {
          if (num[j] == i && num[k] == i + 1 && k < j) {
            p1 = j, p2 = k;
            flag = false;
            break;
          }
        }
      }
    }
    if (p2 == 1 && p1 == n) {
      r[ans] = 2;
      while (num[p2] == num[p2 + 1] - 1) p2++;
      for (int i = 1, j = p2 + 1; i <= n; i++) {
        temp[i] = num[j++];
        if (j > n) j = 1;
      }
      op[ans][0] = p2;
      op[ans][1] = n - p2;
    } else if (p2 == 1 && p1 < n) {
      r[ans] = 3;
      while (num[p2] == num[p2 + 1] - 1) p2++;
      op[ans][0] = p2;
      op[ans][1] = p1 - p2;
      op[ans][2] = n - p1;
      int j = p1 + 1, i = 1;
      while (j <= n) temp[i++] = num[j++];
      j = p2 + 1;
      while (j <= p1) temp[i++] = num[j++];
      j = 1;
      while (j <= p2) temp[i++] = num[j++];
    } else if (p2 > 1 && p1 == n) {
      r[ans] = 3;
      int p22 = p2;
      while (num[p22] == num[p22 + 1] - 1) p22++;
      op[ans][0] = p2 - 1;
      op[ans][1] = p22 - p2 + 1;
      op[ans][2] = n - p22;
      int i = 1, j = p22 + 1;
      while (j <= n) temp[i++] = num[j++];
      j = p2;
      while (j <= p22) temp[i++] = num[j++];
      j = 1;
      while (j < p2) temp[i++] = num[j++];
    } else if (p2 > 1 && p1 < n) {
      r[ans] = 4;
      int p22 = p2;
      while (num[p22] == num[p22 + 1] - 1) p22++;
      op[ans][0] = p2 - 1;
      op[ans][1] = p22 - p2 + 1;
      op[ans][2] = p1 - p22;
      op[ans][3] = n - p1;
      int i = 1, j = p1 + 1;
      while (j <= n) temp[i++] = num[j++];
      j = p22 + 1;
      while (j <= p1) temp[i++] = num[j++];
      j = p2;
      while (j <= p22) temp[i++] = num[j++];
      j = 1;
      while (j < p2) temp[i++] = num[j++];
    }
    for (int i = 1; i <= n; i++) {
      num[i] = temp[i];
    }
  }
  printf("%d\n", ans);
  for (int i = 1; i <= ans; i++) {
    printf("%d ", r[i]);
    for (int j = 0; j < r[i]; j++) printf("%d ", op[i][j]);
    printf("\n");
  }
  return 0;
}