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 t, n;
vector<int> a;
vector<vector<int>> ans;
vector<int> op(vector<int> b, vector<int> blk) {
  int sum = 0;
  vector<int> ret;
  int tmp = 0;
  for (int x : blk) tmp += x;
  assert(tmp == b.size());
  for (int x : blk) {
    sum += x;
    for (int i = b.size() - sum; i < b.size() - sum + x; i++)
      ret.push_back(b[i]);
  }
  return ret;
}
vector<int> comp(vector<int> blk) {
  vector<int> ret;
  for (int x : blk)
    if (x) ret.push_back(x);
  return ret;
}
int main() {
  scanf("%d", &n);
  a.resize(n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    a[i]--;
  }
  while (!is_sorted(a.begin(), a.end())) {
    if (a[0] == 0 || a.back() == n - 1) {
      vector<int> cur;
      int idx = n - 1, pref = 0;
      while (a[pref] == pref) pref++;
      while (a[idx] == idx) {
        cur.push_back(1);
        idx--;
      }
      int nxt = idx;
      while (a[idx] != nxt) idx--;
      cur.push_back(nxt - idx + 1);
      cur.push_back(idx - pref);
      cur.push_back(pref);
      ans.push_back(comp(cur));
      a = op(a, ans.back());
    } else {
      vector<int> cur;
      int p0 = -1;
      for (int i = 0; i < n; i++)
        if (a[i] == 0) p0 = i;
      int flag = 1;
      for (int i = p0, j = 0; i < n && flag; i++, j++)
        if (a[i] != j) flag = 0;
      if (flag) flag = (n - p0);
      cur.push_back(flag);
      int pn = -1;
      for (int i = 0; i < n; i++)
        if (a[i] == flag) pn = i;
      cur.push_back(n - pn - flag);
      int i, j;
      for (i = n - 1, j = 0; j < pn && a[j] == i; i--, j++)
        ;
      cur.push_back(pn - j);
      for (int z = j - 1; z >= 0; z--) cur.push_back(1);
      ans.push_back(comp(cur));
      a = op(a, ans.back());
    }
  }
  printf("%d\n", ans.size());
  for (auto tt : ans) {
    printf("%d", tt.size());
    reverse(tt.begin(), tt.end());
    for (int x : tt) printf(" %d", x);
    puts("");
  }
  return 0;
}