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;
double pi = 2 * acos(0.0);
template <typename T>
ostream &operator<<(ostream &os, const vector<T> &v) {
  os << '{';
  for (const auto &x : v) os << " " << x;
  return os << '}';
}
int n;
vector<int> v, ans[55];
void solve(int kk) {
  int sz = ans[kk].size();
  vector<int> t[sz], v1;
  int ix = 0, cc = 0;
  for (int i = 0; i < sz; i++) {
    for (ix, cc = 0; cc < ans[kk][i]; ix++, cc++) {
      t[i].push_back(v[ix]);
    }
  }
  v.clear();
  for (int i = sz - 1; i >= 0; i--) {
    for (int j = 0; j < t[i].size(); j++) {
      v.push_back(t[i][j]);
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n;
  int in;
  for (int i = 0; i < n; i++) {
    cin >> in;
    v.push_back(in);
  }
  int kk = 0;
  while (true) {
    int ix1 = -1, ix2 = -1;
    map<int, int> mp, ixs;
    for (int i = 0; i < n; i++) {
      mp[v[i]]++;
      ixs[v[i]] = i;
      if (mp[v[i] + 1]) {
        ix1 = ixs[v[i]];
        ix2 = ixs[v[i] + 1];
        mp.clear();
        ixs.clear();
        break;
      }
    }
    if (ix2 == -1 && ix1 == -1) break;
    int x = 1;
    if (ix2) ans[kk].push_back(ix2);
    x = 1;
    for (int j = ix2 + 1; j < ix1; j++) {
      if (v[j] > v[j - 1]) {
        x++;
        ix2 = j;
      } else {
        break;
      }
    }
    ans[kk].push_back(x);
    ans[kk].push_back(ix1 - ix2);
    if (n - 1 - ix1) ans[kk].push_back(n - 1 - ix1);
    solve(kk);
    kk++;
  }
  cout << kk << endl;
  for (int i = 0; i < kk; i++) {
    cout << ans[i].size() << " ";
    for (auto it : ans[i]) cout << it << " ";
    cout << endl;
  }
  return 0;
}