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;
void run_case() {
  int n;
  cin >> n;
  vector<int> c(n);
  vector<int> good(n);
  for (int i = 0; i < n; i++) {
    cin >> c[i];
    good[i] = i + 1;
  }
  vector<vector<int>> ans;
  int cur = 1;
  while (c != good) {
    vector<int> ch;
    vector<int> pos(n + 1);
    for (int i = 0; i < n; i++) pos[c[i]] = i;
    while (cur != n && abs(pos[cur + 1] - pos[cur]) <= 1) {
      cur++;
    }
    if (cur == n) {
      assert(c != good);
      reverse(c.begin(), c.end());
      assert(c == good);
      for (int i = 0; i < n; i++) ch.push_back(1);
      ans.push_back(ch);
      break;
    }
    bool rev1 = pos[cur] < pos[1];
    bool rev2 = pos[cur + 1] < pos[1];
    if (rev1 && rev2) {
      if (pos[cur + 1]) ch.push_back(pos[cur + 1]);
      ch.push_back(pos[cur] - pos[cur + 1]);
      for (int i = 1; i <= cur; i++) ch.push_back(1);
      if (n - 1 - pos[1]) ch.push_back(n - 1 - pos[1]);
    } else if (rev1 && !rev2) {
      if (pos[cur]) ch.push_back(pos[cur]);
      ch.push_back(pos[1] - pos[cur] + 1);
      ch.push_back(pos[cur + 1] - pos[1]);
      if (n - 1 - pos[cur + 1]) ch.push_back(n - 1 - pos[cur + 1]);
    } else if (!rev1 && rev2) {
      if (pos[cur + 1]) ch.push_back(pos[cur + 1]);
      ch.push_back(pos[1] - pos[cur + 1]);
      ch.push_back(pos[cur] - pos[1] + 1);
      if (n - 1 - pos[cur]) ch.push_back(n - 1 - pos[cur]);
    } else {
      if (pos[1]) ch.push_back(pos[1]);
      for (int i = 1; i <= cur; i++) ch.push_back(1);
      ch.push_back(pos[cur + 1] - pos[cur]);
      if (n - 1 - pos[cur + 1]) ch.push_back(n - 1 - pos[cur + 1]);
    }
    ans.push_back(ch);
    vector<vector<int>> temp;
    int prev = 0;
    for (int& i : ch) {
      vector<int> temp2;
      for (int j = prev; j < prev + i; j++) {
        assert(j < n);
        temp2.push_back(c[j]);
      }
      prev = prev + i;
      temp.push_back(temp2);
    }
    reverse(temp.begin(), temp.end());
    int ind = 0;
    for (vector<int>& i : temp) {
      for (int& j : i) {
        c[ind] = j;
        ind++;
      }
    }
  }
  cout << ans.size() << '\n';
  for (vector<int>& i : ans) {
    cout << i.size() << " ";
    for (int& j : i) cout << j << " ";
    cout << "\n";
  }
}
int main() {
  ios::sync_with_stdio(false), cin.tie(0);
  if (fopen("input.in", "r"))
    freopen("input.in", "r", stdin), freopen("output.out", "w", stdout);
  int t;
  t = 1;
  while (t--) {
    run_case();
  }
}