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 long long mod = 1000000007;
const long long INF = 4e18 + 10;
long long n;
long long a[55];
vector<vector<long long> > ans;
long long findpos(long long x) {
  for (long long i = 0; i < n; i++)
    if (a[i] == x) return i;
}
void p1(long long x) {
  long long xx = findpos(x);
  vector<long long> v;
  if (xx > 0) v.push_back(xx);
  if (n - x + 1 - xx > 0) v.push_back(n - x + 1 - xx);
  for (long long i = n - x + 1; i < n; i++) v.push_back(1);
  ans.push_back(v);
  long long b[n];
  long long i = 0;
  for (i = 0; i < x - 1; i++) b[i] = i + 1;
  for (long long j = xx; j <= n - x; j++) {
    b[i] = a[j];
    i++;
  }
  for (long long j = 0; j < xx; j++) {
    b[i] = a[j];
    i++;
  }
  for (long long j = 0; j < n; j++) {
    a[j] = b[j];
  }
}
void p2(long long x) {
  long long xx = findpos(x);
  vector<long long> v;
  for (long long i = 0; i < x - 1; i++) v.push_back(1);
  if (xx + 1 - x + 1 > 0) v.push_back(xx + 1 - x + 1);
  if (n - 1 - xx > 0) v.push_back(n - 1 - xx);
  ans.push_back(v);
  long long b[n];
  long long i = 0;
  for (long long j = xx + 1; j < n; j++) {
    b[i] = a[j];
    i++;
  }
  for (long long j = x - 1; j <= xx; j++) {
    b[i] = a[j];
    i++;
  }
  for (long long j = 0; j < x - 1; j++) {
    b[i] = j + 1;
    i++;
  }
  for (long long j = 0; j < n; j++) {
    a[j] = b[j];
  }
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  cout << fixed << setprecision(10);
  long long t = 1;
  while (t--) {
    cin >> n;
    for (long long i = 0; i < n; i++) cin >> a[i];
    long long x = findpos(1);
    if (n % 2 == 1) {
      if (x == 0) {
      } else {
        vector<long long> v;
        v.push_back(x);
        v.push_back(n - x);
        ans.push_back(v);
        long long b[n];
        long long i = 0;
        for (long long j = x; j < n; j++) {
          b[i] = a[j];
          i++;
        }
        for (long long j = 0; j < x; j++) {
          b[i] = a[j];
          i++;
        }
        for (long long j = 0; j < n; j++) {
          a[j] = b[j];
        }
      }
      for (long long i = 2; i <= n; i++) {
        if (i % 2 == 1)
          p1(i);
        else
          p2(i);
      }
    } else {
      if (x == n - 1) {
      } else {
        vector<long long> v;
        v.push_back(x + 1);
        v.push_back(n - x - 1);
        ans.push_back(v);
        long long b[n];
        long long i = 0;
        for (long long j = x + 1; j < n; j++) {
          b[i] = a[j];
          i++;
        }
        for (long long j = 0; j <= x; j++) {
          b[i] = a[j];
          i++;
        }
        for (long long j = 0; j < n; j++) {
          a[j] = b[j];
        }
      }
      for (long long i = 2; i <= n; i++) {
        if (i % 2 == 1)
          p2(i);
        else
          p1(i);
      }
    }
    cout << ans.size() << endl;
    for (long long i = 0; i < ans.size(); i++) {
      cout << ans[i].size() << " ";
      for (long long j = 0; j < ans[i].size(); j++) {
        cout << ans[i][j] << " ";
      }
      cout << endl;
    }
  }
  return 0;
}