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>
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,avx2,fma")
#pragma GCC optimize("unroll-loops")
#pragma GCC optimize("O3")
#pragma GCC optimize("O2")
#pragma GCC optimize("Og")
#pragma GCC optimize("Ofast")
#pragma GCC optimize("inline")
#pragma GCC optimize("-fgcse")
using namespace std;
using namespace std;
long long MOD = 998245709;
long long exp(long long n, long long e) {
  if (e == 0) return 1;
  long long temp = exp((n * n) % MOD, e / 2);
  return e % 2 == 0 ? temp : (temp * n) % MOD;
}
long long inverse(long long n) { return exp(n, MOD - 2); }
void frac(long long p, long long q) { cout << (p * inverse(q)) % MOD << endl; }
long long ceil_div(long long a, long long b) { return (a + b - 1) / b; }
int n;
vector<vector<long long>> op;
void operation(vector<long long> &arr, int d1, int d2, int d3, int d4) {
  vector<long long> ans(n + 1);
  d4 = d4 - d3;
  d3 = d3 - d2;
  d2 = d2 - d1;
  vector<long long> vec;
  if (d1 > 0) vec.push_back(d1);
  if (d2 > 0) vec.push_back(d2);
  if (d3 > 0) vec.push_back(d3);
  if (d4 > 0) vec.push_back(d4);
  op.push_back(vec);
  for (int i = 1; i <= d4; i++) {
    ans[i] = arr[d1 + d2 + d3 + i];
  }
  for (int i = 1; i <= d3; i++) {
    ans[d4 + i] = arr[d1 + d2 + i];
  }
  for (int i = 1; i <= d2; i++) {
    ans[d4 + d3 + i] = arr[d1 + i];
  }
  for (int i = 1; i <= d1; i++) {
    ans[d4 + d3 + d2 + i] = arr[i];
  }
  arr = ans;
}
void solve() {
  cin >> n;
  vector<long long> arr(n + 1);
  for (int i = 1; i <= n; i++) cin >> arr[i];
  vector<long long> left(n + 1, 0), right(n + 1, 0);
  for (int i = 1; i < n; i++) {
    vector<long long> pos(n + 1, -1);
    for (int j = 1; j <= n; j++) {
      pos[arr[j]] = j;
    }
    int pick = -1;
    for (int j = 2; j <= n; j++) {
      if (pos[j] < pos[j - 1]) {
        pick = j;
        break;
      }
    }
    if (pick == -1) break;
    vector<long long> b;
    int d1 = pos[pick] - 1;
    int d2 = pos[pick - 1] - left[pick - 1] - 1;
    int d3 = pos[pick - 1];
    int d4 = n;
    operation(arr, d1, d2, d3, d4);
    for (int j = 0; j <= right[pick]; j++) {
      left[pick + j] += left[pick - 1] + 1;
    }
    for (int j = 0; j <= left[pick - 1]; j++) {
      right[pick - 1 - j] += right[pick] + 1;
    }
  }
  cout << ((long long)(op).size()) << endl;
  for (int i = 0; i < ((long long)(op).size()); i++) {
    cout << ((long long)(op[i]).size()) << " ";
    for (int j = 0; j < ((long long)(op[i]).size()); j++)
      cout << op[i][j] << " ";
    cout << endl;
  }
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t = 1;
  for (long long i = 1; i <= t; i++) {
    solve();
  }
  return 0;
}