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 comment(linker, "/STACK:9759095000")
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#pragma GCC target("sse,sse2,sse3,ssse3,abm,mmx,tune=native")
using namespace std;
long long gcd(long long i, long long j) {
  if (j == 0)
    return i;
  else
    return gcd(j, i % j);
}
template <typename T>
inline T getint() {
  T val = 0;
  char c;
  bool neg = false;
  while ((c = getchar()) && !(c >= '0' && c <= '9')) {
    neg |= c == '-';
  }
  do {
    val = (val * 10) + c - '0';
  } while ((c = getchar()) && (c >= '0' && c <= '9'));
  return val * (neg ? -1 : 1);
}
const long long INF = 1e9;
const int mod = 1000000007;
const long double eps = 1e-9, pi = acos(-1);
const int maxN = 1000100, maxT = 2001;
void file_inc() {
  freopen("input.txt", "r", stdin);
  freopen("output.txt", "w", stdout);
}
bool valid(vector<int> cc) {
  for (int i = 0; i < cc.size(); ++i) {
    if (i != cc[i]) return false;
  }
  return true;
}
vector<int> build_idx(vector<int> c) {
  vector<int> res(c.size());
  for (int i = 0; i < c.size(); ++i) {
    res[c[i]] = i;
  }
  return res;
}
int32_t main() {
  cin.tie(0);
  cout.tie(0);
  ios_base::sync_with_stdio(false);
  cout.precision(15);
  int n;
  cin >> n;
  vector<int> c(n), ID(n);
  vector<vector<int>> q;
  for (auto& x : c) {
    cin >> x;
    --x;
  }
  while (!valid(c)) {
    ID = build_idx(c);
    int i1 = 0, i2 = 0;
    for (int i = 1; i < ID.size(); ++i) {
      if (ID[i] < ID[i - 1]) {
        i1 = ID[i - 1];
        i2 = ID[i];
        break;
      }
    }
    int j2 = i2 + 1;
    while (j2 < n && c[j2 - 1] == c[j2] - 1) ++j2;
    vector<int> cc;
    q.push_back({});
    if (i2 != 0) {
      q.back().push_back(i2);
    }
    q.back().push_back(j2 - i2);
    q.back().push_back(i1 - j2 + 1);
    if (i1 + 1 != n) {
      q.back().push_back(n - i1 - 1);
    }
    if (i1 + 1 != n) {
      for (int i = i1 + 1; i < n; ++i) cc.push_back(c[i]);
    }
    for (int i = j2; i <= i1; ++i) cc.push_back(c[i]);
    for (int i = i2; i < j2; ++i) cc.push_back(c[i]);
    for (int i = 0; i < i2; ++i) cc.push_back(c[i]);
    c = cc;
  }
  cout << q.size() << "\n";
  for (auto x : q) {
    cout << x.size() << " ";
    for (auto y : x) cout << y << " ";
    cout << "\n";
  }
  return 0;
}