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;
priority_queue<long long, std::vector<long long>, std::greater<long long> > pqs;
priority_queue<long long> pq;
queue<long long> qq;
stack<long long> sta;
set<long long> st;
set<long long>::iterator P;
pair<long long, long long> pp;
vector<long long> v[55], vv[55];
map<long long, long long> mp, pm;
long long i, j, n, m, t, k, a, b, c, x, mx, mn;
string s, s1;
char z;
int main() {
  ios_base::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
  int T = 1;
  while (T--) {
    i = j = n = m = t = k = a = b = c = x = mx = mn = 0;
    cin >> n;
    long long A[n + 5];
    for (i = 0; i < n; i++) cin >> A[i];
    long long index = 0;
    while (k == 0) {
      k = 1, j = 0;
      for (i = 0; i < n; i++) {
        if (a % 2 == 0) {
          v[j].push_back(A[i]);
          while (i < n - 1 && A[i] < A[i + 1]) v[j].push_back(A[i + 1]), i++;
          j++;
        } else {
          v[j].push_back(A[i]);
          while (i < n - 1 && A[i] > A[i + 1]) v[j].push_back(A[i + 1]), i++;
          j++;
        }
      }
      if (j == 1) return cout << 0 << endl, 0;
      i = 0;
      for (long long ii = j; ii >= 0; ii--)
        for (k = 0; k < v[ii].size(); k++) A[i] = v[ii][k], i++;
      vv[index].push_back(j);
      for (i = 0; i < j; i++) vv[index].push_back(v[i].size());
      index++;
      for (i = 0; i < 53; i++) v[i].clear();
      for (i = 0; i < n; i++)
        if (A[i] != i + 1) k = 0;
      a++;
    }
    cout << index << endl;
    for (i = 0; i < index; i++) {
      for (j = 0; j < vv[i].size(); j++) cout << vv[i][j] << " ";
      cout << endl;
    }
  }
  return 0;
}