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;
vector<int> get_pos(const vector<int>& deck) {
  int n = deck.size();
  vector<int> pos(n);
  for (int i = 0; i < n; i++) pos[deck[i]] = i;
  return pos;
}
vector<int> get_new_deck(const vector<int>& deck, const vector<int>& v) {
  int n = deck.size();
  vector<int> ndeck;
  int s = 0;
  for (auto l : v) s += l;
  assert(s == n);
  int sp = n;
  for (int k = v.size() - 1; k >= 0; k--) {
    int l = v[k];
    sp -= l;
    for (int i = sp, j = 0; j < l; i++, j++) ndeck.push_back(deck[i]);
  }
  return ndeck;
}
int get_level(const vector<int>& deck) {
  int n = deck.size();
  auto pos = get_pos(deck);
  int p = pos[0];
  int lvf = 1;
  for (int i = p + 1; i < n; i++) {
    if (deck[i] == lvf)
      lvf++;
    else
      break;
  }
  int lvb = 1;
  for (int i = p - 1; i >= 0; i--) {
    if (deck[i] == lvb)
      lvb++;
    else
      break;
  }
  return max(lvf, lvb);
}
int main() {
  int n;
  vector<vector<int>> ans;
  cin >> n;
  vector<int> deck(n);
  for (int i = 0; i < n; i++) {
    cin >> deck[i];
    deck[i]--;
  }
  int lv = get_level(deck);
  auto pos = get_pos(deck);
  while (lv < n) {
    int ct = lv - 1;
    int nt = lv;
    bool isf = (ct == 0 || pos[ct - 1] < pos[ct]);
    vector<int> v;
    if (isf) {
      if (pos[nt] > pos[ct]) {
        if (pos[nt] != n - 1) v.push_back(n - 1 - pos[nt]);
        v.push_back(pos[nt] - pos[ct]);
        for (int i = 0; i < lv; i++) v.push_back(1);
        if (pos[0] != 0) v.push_back(pos[0]);
        reverse(begin(v), end(v));
        ans.push_back(v);
      } else {
        if (pos[ct] != n - 1) v.push_back(n - 1 - pos[ct]);
        v.push_back(lv);
        v.push_back(pos[0] - pos[nt]);
        if (pos[nt] != 0) v.push_back(pos[nt]);
        reverse(begin(v), end(v));
        ans.push_back(v);
      }
    } else {
      if (pos[nt] > pos[ct]) {
        if (pos[nt] != n - 1) v.push_back(n - 1 - pos[nt]);
        v.push_back(1);
        v.push_back(pos[nt] - pos[ct]);
        if (pos[ct] != 0) v.push_back(pos[ct]);
        reverse(begin(v), end(v));
        ans.push_back(v);
      } else {
        if (pos[0] != n - 1) v.push_back(n - 1 - pos[0]);
        for (int i = 0; i < lv; i++) v.push_back(1);
        v.push_back(pos[ct] - pos[nt]);
        if (pos[nt] != 0) v.push_back(pos[nt]);
        reverse(begin(v), end(v));
        ans.push_back(v);
      }
    }
    deck = get_new_deck(deck, v);
    lv = get_level(deck);
    pos = get_pos(deck);
  }
  if (pos[0] != 0) {
    vector<int> v(n, 1);
    ans.push_back(v);
  }
  cout << ans.size() << endl;
  for (auto& v : ans) {
    cout << v.size();
    for (auto val : v) cout << " " << val;
    cout << endl;
  }
  return 0;
}