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;
const int INF = (int)1e9 + 41;
inline void mxe(int &a, int b) {
  if (b > a) a = b;
}
void kek(vector<int> &a, const vector<int> &v) {
  int cur = 0;
  for (int x : v) {
    reverse(a.begin() + cur, a.begin() + cur + x);
    cur += x;
  }
  reverse(a.begin(), a.end());
}
void solve() {
  int n;
  cin >> n;
  vector<int> a(n);
  for (int &x : a) {
    cin >> x;
    x--;
  }
  auto fnd = [&](int x) {
    for (int i = 0; i < n; i++)
      if (a[i] == x) return i;
    exit(11);
  };
  int p0 = fnd(0);
  bool isL = true;
  if (n % 2 == 1) {
    if (p0 == 0)
      cout << n - 1 << "\n";
    else {
      cout << n << "\n"
           << "2 " << p0 << " " << n - p0 << "\n";
      kek(a, {p0, n - p0});
    }
  } else {
    isL = false;
    if (p0 + 1 == n)
      cout << n - 1 << "\n";
    else {
      cout << n << "\n"
           << "2 " << p0 + 1 << " " << n - p0 - 1 << "\n";
      kek(a, {p0 + 1, n - p0 - 1});
    }
  }
  auto print = [&](const vector<int> &v) {
    cout << v.size();
    for (int i : v) cout << " " << i;
    cout << "\n";
  };
  for (int i = 1; i < n; i++) {
    int pos = fnd(i);
    vector<int> v;
    for (int j = 0; j < i; j++) v.push_back(1);
    if (isL) {
      v.push_back(pos - i + 1);
      if (pos + 1 < n) v.push_back(n - pos - 1);
    } else {
      v.push_back(n - pos - i);
      if (pos > 0) v.push_back(pos);
      reverse(v.begin(), v.end());
    }
    print(v);
    kek(a, v);
    isL = !isL;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  solve();
}