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;
int n;
vector<int> a, b;
vector<vector<array<int, 2>>> ans;
void doswap(vector<array<int, 2>> op) {
  b = vector<int>();
  reverse((op).begin(), (op).end());
  for (int i = (0); i < (((int)(op.size()))); ++i) {
    int l = op[i][0], r = op[i][1];
    for (int j = (l); j < (r + 1); ++j) b.push_back(a[j]);
  }
  swap(a, b);
}
bool check() {
  for (int i = (0); i < (n - 1); ++i)
    if (a[i] > a[i + 1]) return 0;
  return 1;
}
bool recheck() {
  for (int i = (0); i < (n - 1); ++i)
    if (a[i] < a[i + 1]) return 0;
  return 1;
}
int main() {
  scanf("%d", &n);
  a.resize(n);
  for (int i = (0); i < (n); ++i) scanf("%d", &a[i]);
  int cnt = 0;
  while (!check()) {
    cnt++;
    assert(cnt < n + 1);
    for (int i = (1); i < (n); ++i) {
      int p1l, p1r, p2l, p2r;
      p1r = find((a).begin(), (a).end(), i) - a.begin();
      p1l = p1r;
      p2l = find((a).begin(), (a).end(), i + 1) - a.begin();
      p2r = p2l;
      if (p1r > p2l) {
        while (p2r + 1 < n && a[p2r + 1] == a[p2r] + 1) p2r++;
        while (p1l - 1 > -1 && a[p1l - 1] == a[p1l] - 1) p1l--;
        vector<array<int, 2>> op;
        if (p2l != 0) {
          op.push_back({0, p2l - 1});
        }
        op.push_back({p2l, p2r});
        op.push_back({p2r + 1, p1r});
        if (p1r + 1 < n) {
          op.push_back({p1r + 1, n - 1});
        }
        ans.push_back(op);
        doswap(op);
        break;
      }
    }
  }
  printf("%d\n", ((int)(ans.size())));
  for (auto &sb : ans) {
    printf("%d ", ((int)(sb.size())));
    for (int i = (0); i < (((int)(sb.size()))); ++i)
      printf("%d%c", sb[i][1] - sb[i][0] + 1,
             " \n"[i + 1 == ((int)(sb.size()))]);
  }
  return 0;
}