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;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
int n, a[60], t = 0;
vector<int> v;
void solve(vector<int> vv) {
  cout << vv.size();
  for (auto i : vv) cout << " " << i;
  cout << endl;
  vector<vector<int>> te;
  int cnt = 1;
  for (auto i : vv) {
    vector<int> tte;
    for (int j = 1; j <= i; j++) tte.push_back(a[cnt++]);
    te.push_back(tte);
  }
  reverse(te.begin(), te.end());
  cnt = 1;
  for (auto i : te)
    for (auto j : i) a[cnt++] = j;
}
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  int ans = (n - 1) / 2 * 2 + 1;
  if (a[1] == 1) ans--;
  cout << ans << endl;
  for (int i = 1; i <= (n - 1) / 2 * 2 + 1; i++) {
    t ^= 1;
    v.clear();
    if (t) {
      if (i == 1 && a[1] == 1) continue;
      if (a[1] == i)
        v.push_back(n - i + 1);
      else {
        for (int j = 1; j <= n; j++) {
          if (a[j] == i) {
            v.push_back(j - 1);
            v.push_back(n - i - j + 2);
          }
        }
      }
      for (int j = 1; j < i; j++) v.push_back(1);
    } else {
      for (int j = 1; j < i; j++) v.push_back(1);
      if (a[n] == i)
        v.push_back(n - i + 1);
      else {
        for (int j = 1; j <= n; j++) {
          if (a[j] == i) {
            v.push_back(j - i + 1);
            v.push_back(n - j);
          }
        }
      }
    }
    solve(v);
  }
  return 0;
}