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;
void err(istream_iterator<string> it) {}
template <typename T, typename... Args>
void err(istream_iterator<string> it, T a, Args... args) {
  cerr << *it << " = " << a << endl;
  err(++it, args...);
}
inline int add(int a, int b) {
  a += b;
  return a >= 1000000007 ? a - 1000000007 : a;
}
inline int sub(int a, int b) {
  a -= b;
  return a < 0 ? a + 1000000007 : a;
}
inline int mul(int a, int b) { return (long long int)a * b % 1000000007; }
int Set(int N, int pos) { return N = N | (1 << pos); }
int Reset(int N, int pos) { return N = N & ~(1 << pos); }
bool Cheek(int N, int pos) { return (bool)(N & (1 << pos)); }
int ara[60];
void Solve(int t) {
  int i, j, k, n, pos1 = -1, pos2 = -1, pos3 = -1, pos4 = -1;
  deque<int> dq1, dq2, dq3, dq4;
  vector<vector<int> > ans;
  scanf("%d", &(n));
  for (i = 1; i <= n; i++) scanf("%d", &(ara[i]));
  int ncopy = n;
  while (ncopy--) {
    for (i = 1; i < n; i++) {
      pos1 = pos2 = -1;
      dq1.clear();
      dq2.clear();
      dq3.clear();
      dq4.clear();
      for (j = i + 1; j <= n; j++) {
        if (ara[i] == 1 + ara[j]) {
          pos1 = i;
          pos2 = j;
          break;
        }
      }
      if (pos1 == -1 && pos2 == -1) continue;
      for (k = pos1; k < pos2; k++) {
        if (ara[k] > ara[k + 1]) {
          pos3 = k;
          break;
        }
      }
      for (j = 1; j < pos1; j++) dq1.push_back(ara[j]);
      for (j = pos1; j <= pos3; j++) dq2.push_back(ara[j]);
      for (j = pos3 + 1; j <= pos2; j++) dq3.push_back(ara[j]);
      for (j = pos2 + 1; j <= n; j++) dq4.push_back(ara[j]);
      vector<int> temp;
      int pos = 1;
      if (dq4.size() > 0) {
        temp.push_back(dq4.size());
        while (!dq4.empty()) {
          ara[pos++] = dq4.front();
          dq4.pop_front();
        }
      }
      if (dq3.size() > 0) {
        temp.push_back(dq3.size());
        while (!dq3.empty()) {
          ara[pos++] = dq3.front();
          dq3.pop_front();
        }
      }
      if (dq2.size() > 0) {
        temp.push_back(dq2.size());
        while (!dq2.empty()) {
          ara[pos++] = dq2.front();
          dq2.pop_front();
        }
      }
      if (dq1.size() > 0) {
        temp.push_back(dq1.size());
        while (!dq1.empty()) {
          ara[pos++] = dq1.front();
          dq1.pop_front();
        }
      }
      reverse(temp.begin(), temp.end());
      ;
      if (!temp.empty()) {
        ans.push_back(temp);
        break;
      }
    }
  }
  cout << ans.size() << endl;
  for (i = 0; i < (int)ans.size(); i++) {
    cout << ans[i].size();
    for (auto x : ans[i]) cout << ' ' << x;
    cout << endl;
  }
}
int main() {
  int t, T;
  T = 1;
  for (__typeof(t) t = 1; t <= T; t++) {
    Solve(t);
  }
  return 0;
}