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 c[Maxn], nc[Maxn];
vector<vector<int> > res;
void Op(const vector<int> &D) {
  res.push_back(D);
  int cur = n + 1;
  int pnt = 1;
  for (int i = 0; i < D.size(); i++) {
    cur -= D[i];
    for (int j = 0; j < D[i]; j++) nc[cur + j] = c[pnt++];
  }
  for (int i = 1; i <= n; i++) c[i] = nc[i];
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &c[i]);
  while (true) {
    vector<pair<int, int> > ints;
    vector<int> D;
    for (int i = 1, j; i <= n; i = j) {
      j = i + 1;
      while (j <= n && c[j] == c[j - 1] + 1) j++;
      ints.push_back(pair<int, int>(i, j - 1));
    }
    bool was = false;
    for (int i = 0; i < ints.size() && !was; i++)
      for (int j = i + 1; j < ints.size() && !was; j++)
        if (c[ints[i].first] == c[ints[j].second] + 1) {
          was = true;
          if (ints[i].first > 1) D.push_back(ints[i].first - 1);
          D.push_back(ints[i].second - ints[i].first + 1);
          D.push_back(ints[j].second - ints[i].second);
          if (ints[j].second < n) D.push_back(n - ints[j].second);
          Op(D);
        }
    if (!was) break;
  }
  printf("%d\n", int(res.size()));
  for (int i = 0; i < res.size(); i++) {
    printf("%d", int(res[i].size()));
    for (int j = 0; j < res[i].size(); j++) printf(" %d", res[i][j]);
    printf("\n");
  }
  return 0;
}