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>
#pragma GCC optimize("Ofast")
using namespace std;
const int maxN = 60;
int T[maxN], pos[maxN], n;
void sym(vector<int> p) {
  reverse(T + 1, T + n + 1);
  reverse((p).begin(), (p).end());
  int i = 1;
  for (int d : p) {
    reverse(T + i, T + i + d);
    i += d;
  }
  for (int i = (1); i < (n + 1); i++) pos[T[i]] = i;
}
void solve() {
  scanf("%d", &n);
  vector<vector<int> > ans;
  for (int i = (1); i < (n + 1); i++) scanf("%d", T + i), pos[T[i]] = i;
  while (true) {
    int k = 0;
    for (int i = (1); i < (n); i++) {
      if (pos[i + 1] < pos[i]) {
        k = i;
        break;
      }
    }
    if (k == 0) break;
    vector<int> temp;
    int i = pos[k], j = pos[k + 1];
    if (j != 1) temp.push_back(j - 1);
    int s = j + 1;
    while (T[s] == T[s - 1] + 1) s++;
    temp.push_back(s - j);
    temp.push_back(i - s + 1);
    if (i != n) temp.push_back(n - i);
    sym(temp);
    ans.push_back(temp);
    0;
    0;
    0;
    0;
    0;
  }
  printf("%d\n", ((int)(ans).size()));
  for (auto V : ans) {
    printf("%d ", ((int)(V).size()));
    for (int v : V) printf("%d ", v);
    printf("\n");
  }
}
int main() {
  int t = 1;
  for (int tid = (1); tid < (t + 1); tid++) {
    solve();
  }
  return 0;
}