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;
int qwe, qwr;
const long long MOD = 1e9 + 7, MAX = 53;
const int INF = 2e9;
int n, k, tt;
int I = -1, J = -1;
string s;
vector<int> a;
vector<int> v[MAX];
bool ver(int i, int j) {
  for (int i = 1; i < n; i++)
    if (a[tt - 1] == i and a[tt] == j) {
      k = tt - 1;
      return true;
    }
  k = -1;
  return false;
}
bool see(int i, int j) {
  for (int lx = 0; lx < n; lx++) {
    if (a[lx] == i) I = lx;
    if (a[lx] == j) J = lx;
  }
  return (J > I);
}
void simulation(int l, vector<int> vk) {
  vector<int> v1;
  int v2[55];
  v2[0] = 0;
  for (int i = 0; i < l; i++) {
    v2[i + 1] = v2[i] + vk[i];
  }
  for (int i = l - 1; i > -1; i--) {
    for (int j = 0; j < vk[i]; j++) v1.push_back(a[v2[i] + j]);
  }
  a.clear();
  for (auto it : v1) a.push_back(it);
}
void read(vector<int>& a) {
  cin >> qwe;
  a.push_back(qwe);
}
void read2(vector<pair<int, int>>& a) {
  cin >> qwe >> qwr;
  a.push_back({qwe, qwr});
}
void solve() {
  cin >> n;
  for (int i = 0; i < n; i++) read(a);
  int num = 0;
  while (true) {
    int j = 0;
    while (a[j] == j + 1 and j < n) j++;
    if (j == n) break;
    j--;
    if (see(j + 2, j + 3)) {
      v[num].push_back(j + 1);
      v[num].push_back(I - j);
      v[num].push_back(J - I);
      v[num].push_back(n - J - 1);
      simulation(4, v[num]);
      num++;
      v[num].push_back(n - I - 2);
      v[num].push_back(I - j);
      v[num].push_back(1);
      v[num].push_back(j + 1);
      simulation(4, v[num]);
      num++;
    } else if (see(j + 3, j + 2)) {
      if (J != n - 1 or j != -1) {
        v[num].push_back(j + 1);
        v[num].push_back(J - j);
        v[num].push_back(n - 1 - J);
        simulation(3, v[num]);
        num++;
      }
      v[num].push_back(n - 2 - J + I - j);
      v[num].push_back(J - I);
      v[num].push_back(1);
      v[num].push_back(j + 1);
      simulation(4, v[num]);
      num++;
    }
  }
  cout << num << '\n';
  for (int i = 0; i < num; i++) {
    int ga = 0;
    for (auto it : v[i])
      if (it != 0) ga++;
    cout << ga << ' ';
    for (auto it : v[i])
      if (it != 0) cout << it << ' ';
    cout << '\n';
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  solve();
  cout << '\n';
  return 0;
}