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 solve();
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  bool HAST = 0;
  if (HAST) {
    int t;
    cin >> t;
    for (int tt = 0, _tt = (t); tt < _tt; ++tt) {
      solve();
      cout << '\n';
    }
  } else {
    solve();
  }
}
void applymove(vector<int>& a, vector<int>& move) {
  vector<int> b(int((a).size()));
  if (int((move).size()) == 2) {
    for (int i = 0, _i = (move[1]); i < _i; ++i) {
      b[i] = a[i + move[0]];
    }
    for (int i = 0, _i = (move[0]); i < _i; ++i) {
      b[i + move[1]] = a[i];
    }
  } else if (int((move).size()) == 3) {
    for (int i = 0, _i = (move[2]); i < _i; ++i) {
      b[i] = a[i + move[0] + move[1]];
    }
    for (int i = 0, _i = (move[1]); i < _i; ++i) {
      b[i + move[2]] = a[i + move[0]];
    }
    for (int i = 0, _i = (move[0]); i < _i; ++i) {
      b[i + move[2] + move[1]] = a[i];
    }
  } else {
    for (int i = 0, _i = (move[3]); i < _i; ++i) {
      b[i] = a[i + move[0] + move[1] + move[2]];
    }
    for (int i = 0, _i = (move[2]); i < _i; ++i) {
      b[i + move[3]] = a[i + move[0] + move[1]];
    }
    for (int i = 0, _i = (move[1]); i < _i; ++i) {
      b[i + move[3] + move[2]] = a[i + move[0]];
    }
    for (int i = 0, _i = (move[0]); i < _i; ++i) {
      b[i + move[3] + move[2] + move[1]] = a[i];
    }
  }
  for (int i = 0, _i = (int((a).size())); i < _i; ++i) a[i] = b[i];
}
bool iscomplete(vector<int>& a) {
  for (int i = 0, _i = (int((a).size())); i < _i; ++i) {
    if (a[i] != i + 1) return 0;
  }
  return 1;
}
void solve() {
  int n;
  cin >> n;
  vector<int> a(n);
  for (int i = 0, _i = (n); i < _i; ++i) cin >> a[i];
  bool complete = 0;
  vector<vector<int> > moves;
  if (n == 1) {
    cout << 0;
    return;
  }
  complete = iscomplete(a);
  while (!complete) {
    int notouch = 0;
    while (a[notouch] == notouch + 1) notouch++;
    vector<int> move;
    if (notouch > 0) move.push_back(notouch);
    int firstout = notouch;
    while (a[firstout + 1] == a[firstout] + 1) firstout++;
    move.push_back(firstout + 1 - notouch);
    int previ = firstout;
    while (a[previ] != a[notouch] - 1) previ++;
    move.push_back(previ - firstout);
    if (previ < n - 1) move.push_back(n - previ - 1);
    applymove(a, move);
    complete = iscomplete(a);
    moves.push_back(move);
  }
  cout << int((moves).size()) << '\n';
  for (int i = 0, _i = (int((moves).size())); i < _i; ++i) {
    cout << int((moves[i]).size()) << ' ';
    for (int j = 0, _j = (int((moves[i]).size())); j < _j; ++j)
      cout << moves[i][j] << ' ';
    cout << '\n';
  }
}