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 maxn = 55;
int n;
int a[maxn];
int b[maxn];
vector<vector<pair<int, int> > > ans;
bool check() {
  for (int i = 1; i <= n; ++i)
    if (a[i] != i) return 0;
  return 1;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%d", a + i);
  a[0] = -1e9;
  while (!check()) {
    int flag = false;
    for (int i = 1; i <= n; ++i) {
      int x = a[i];
      for (int j = i; j <= n; ++j) {
        int y = a[j];
        if (x == y + 1) {
          vector<pair<int, int> > tmp;
          int tl = i, tr = j;
          while (a[tl + 1] == a[tl] + 1) ++tl;
          while (a[tr - 1] == a[tr] - 1) --tr;
          memcpy(b, a, sizeof(b));
          int p = 0;
          if (i != 1) tmp.push_back(pair<int, int>(1, i - 1));
          tmp.push_back(pair<int, int>(i, tr - 1));
          tmp.push_back(pair<int, int>(tr, j));
          if (j != n) tmp.push_back(pair<int, int>(j + 1, n));
          ans.push_back(tmp);
          for (int k = tmp.size() - 1; k >= 0; --k) {
            int l = tmp[k].first, r = tmp[k].second;
            for (int z = l; z <= r; ++z) a[++p] = b[z];
          }
          flag = true;
        }
        if (flag) break;
      }
      if (flag) break;
    }
  }
  cout << ans.size() << '\n';
  for (auto i : ans) {
    cout << i.size();
    for (auto j : i) printf(" %d", j.second - j.first + 1);
    puts("");
  }
}