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;
const int maxn = 5e5 + 5;
const int inf = 0x3f3f3f3f;
int cnt, m, n, a[maxn], b[maxn], c[maxn];
vector<int> op[maxn];
void cl() {
  cnt = 0;
  for (int i = 1; i <= n;) {
    int j = i + 1;
    while (j <= n && a[j] == a[j - 1] + 1) j++;
    b[++cnt] = j - i;
    i = j;
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
  }
  int f = 0;
  for (int i = 1; i <= n; i++) {
    if (a[i] != n - i + 1) f = 1;
  }
  if (f == 0 && n >= 2) {
    printf("1\n");
    printf("%d ", n);
    for (int i = 1; i <= n; i++) printf("%d ", 1);
    return 0;
  }
  cl();
  while (cnt != 1) {
    int x = a[1];
    int y = a[1] - 1, p = -1, f = 0;
    if (y == 0) {
      y = a[b[1] + 1] - 1;
      f = 1;
    }
    for (int i = 1; i <= n; i++) {
      if (a[i] == y) {
        p = i;
        break;
      }
    }
    int k = 0;
    if (f) {
      for (int i = p + 1; i <= n; i++) c[++k] = a[i];
      c[++k] = a[p];
      for (int i = b[1] + 1; i < p; i++) c[++k] = a[i];
      for (int i = 1; i <= b[1]; i++) c[++k] = a[i];
      for (int i = 1; i <= n; i++) a[i] = c[i];
      op[++m].push_back(b[1]);
      op[m].push_back(p - 1 - b[1] + 1 - 1);
      op[m].push_back(1);
      if (p + 1 <= n) op[m].push_back(n - p);
      cl();
      continue;
    }
    for (int i = p + 1; i <= n; i++) c[++k] = a[i];
    for (int i = b[1] + 1; i <= p; i++) c[++k] = a[i];
    for (int i = 1; i <= b[1]; i++) c[++k] = a[i];
    for (int i = 1; i <= n; i++) a[i] = c[i];
    op[++m].push_back(b[1]);
    op[m].push_back(p - b[1]);
    if (p + 1 <= n) op[m].push_back(n - p);
    cl();
  }
  assert(m <= n);
  printf("%d\n", m);
  for (int i = 1; i <= m; i++) {
    printf("%d ", op[i].size());
    for (int j : op[i]) printf("%d ", j);
    printf("\n");
  }
}