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;
const int mod = 1000000007;
const vector<int> mods = {998244353, 1000000007, 1000000009};
const long long inf = 3e18;
const double pi = acos(-1.0);
const double eps = 0.0000001;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
template <class... K>
using umap = unordered_map<K...>;
template <class... K>
using uset = unordered_set<K...>;
const int N = 100;
long long n, a[N], idx[N];
bool rev;
vector<vector<long long> > res;
void hp(int i, int j, int d = 0) {
  if (i > j) return;
  if (!d) res.back().push_back(j - i + 1);
  while (i < j) {
    swap(idx[a[i]], idx[a[j]]);
    swap(a[i], a[j]);
    i++;
    j--;
  }
}
int main() {
  cin.sync_with_stdio(0);
  cin.tie(0);
  cout.precision(13);
  cin >> n;
  for (long long i = (0); i < (n); ++i) {
    cin >> a[i];
    a[i]--;
    idx[a[i]] = i;
  }
  for (long long i = (0); i < (n - 1); ++i) {
    res.push_back({});
    if (idx[i] == i) {
      for (long long j = (0); j < (n); ++j) hp(j, j);
      continue;
    }
    if (i) {
      long long pr = idx[i - 1];
      long long cur = idx[i];
      for (long long k = (0); k < (pr + 1); ++k) {
        hp(k, k);
      }
      hp(pr + 1, cur);
      hp(cur + 1, n - 1);
    }
    if (!i) {
      long long cur = idx[i];
      if (cur == n - 1) {
        rev = 1;
        hp(0, n - 1, 1);
      } else {
        hp(0, cur);
        hp(cur + 1, n - 1);
      }
    }
  }
  if (((int)(res).size()) % 2 == 1) {
    res.push_back({});
    for (long long i = (0); i < (n); ++i) hp(i, i);
  }
  cout << ((int)(res).size()) - rev << "\n";
  for (long long i = (0); i < (((int)(res).size())); ++i) {
    if (!((int)(res[i]).size())) continue;
    cout << ((int)(res[i]).size()) << " ";
    for (long long j = (0); j < (((int)(res[i]).size())); ++j)
      cout << res[i][i % 2 == 1 ? ((int)(res[i]).size()) - 1 - j : j] << " ";
    cout << "\n";
  }
  return 0;
}