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;
inline long long read() {
  long long x = 0;
  int f = 0;
  char ch = getchar();
  while (ch < '0' || ch > '9') f |= (ch == '-'), ch = getchar();
  while (ch >= '0' && ch <= '9')
    x = (x << 3) + (x << 1) + ch - '0', ch = getchar();
  return x = f ? -x : x;
}
int lowbit(int x) { return x & (-x); }
template <class T>
T big(const T &a1, const T &a2) {
  return a1 > a2 ? a1 : a2;
}
template <class T>
T sml(const T &a1, const T &a2) {
  return a1 < a2 ? a1 : a2;
}
template <typename T, typename... R>
T big(const T &f, const R &...r) {
  return big(f, big(r...));
}
template <typename T, typename... R>
T sml(const T &f, const R &...r) {
  return sml(f, sml(r...));
}
void debug_out() { cout << '\n'; }
template <typename T, typename... R>
void debug_out(const T &f, const R &...r) {
  cout << f << " ";
  debug_out(r...);
}
const int INF = 0x3f3f3f3f;
const int mod = 1e6 + 3;
const int MXN = 5e5 + 5;
const int MXE = 1e6 + 5;
int n, m;
int main() {
  n = read();
  vector<int> ar(n), pos(n + 1);
  for (register int i = 0; i < n; ++i) ar[i] = read();
  vector<vector<int>> ans;
  while (!is_sorted((ar).begin(), (ar).end()) && (int)ans.size() < n) {
    vector<int> vs;
    int one = -1, two = -1;
    for (register int i = 0; i < n; ++i) {
      pos[ar[i]] = i;
    }
    for (register int i = 1; i < n; ++i) {
      if (pos[i] > pos[i + 1]) {
        one = pos[i + 1], two = pos[i];
        break;
      }
    }
    int x = one, len = 1;
    while (ar[x] == ar[x + 1] - 1) ++x, ++len;
    if (one) vs.emplace_back(one);
    vs.emplace_back(len);
    vs.emplace_back(two - one - len + 1);
    if (n - 1 - two) vs.emplace_back(n - 1 - two);
    while (vs.back() == 0) vs.pop_back();
    ans.emplace_back(vs);
    reverse((vs).begin(), (vs).end());
    vector<int> tmp, ret;
    int ti = 0, tj = n;
    for (int x : vs) {
      while (x--) {
        tmp.emplace_back(ar[--tj]);
      }
      reverse((tmp).begin(), (tmp).end());
      for (int y : tmp) ret.emplace_back(y);
      tmp.clear();
    }
    ar = ret;
  }
  printf("%d\n", ans.size());
  for (auto x : ans) {
    printf("%d ", x.size());
    for (int y : x) {
      printf("%d ", y);
    }
    printf("\n");
  }
  return 0;
}