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 double pi = acos(-1.0);
const double eps = 1e-8;
const int mxn = 100033;
const int mod = 1e9 + 7;
inline int read() {
  int x = 0, f = 1;
  char c = getchar();
  while (!isdigit(c)) f = c == '-' ? -1 : 1, c = getchar();
  while (isdigit(c)) x = x * 10 + c - '0', c = getchar();
  return x * f;
}
inline long long gcd(long long a, long long b) {
  return b == 0 ? a : gcd(b, a % b);
}
long long ksm(long long a, long long b, long long mod) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = (ans * a) % mod;
    a = (a * a) % mod;
    b >>= 1;
  }
  return ans;
}
long long inv2(long long a, long long mod) { return ksm(a, mod - 2, mod); }
void exgcd(long long a, long long b, long long &x, long long &y, long long &d) {
  if (!b) {
    d = a;
    x = 1;
    y = 0;
  } else {
    exgcd(b, a % b, y, x, d);
    y -= x * (a / b);
  }
}
int dx[] = {0, 1, 0, -1};
int dy[] = {1, 0, -1, 0};
int n;
int main() {
  std::ios::sync_with_stdio(false);
  std::cin.tie(0);
  std::cout.tie(0);
  ;
  cin >> n;
  vector<int> a(n);
  for (int i = 0; i < n; ++i) cin >> a[i], a[i]--;
  vector<vector<int> > res;
  while (!is_sorted(a.begin(), a.end())) {
    vector<int> pos(n);
    for (int i = 0; i < n; ++i) pos[a[i]] = i;
    int x = 0;
    while (x < n && pos[x] < pos[x + 1]) ++x;
    int y = x + 1;
    while (y + 1 < n && pos[y + 1] == pos[y] + 1) ++y;
    vector<int> v;
    if (pos[x + 1]) v.push_back(pos[x + 1]);
    v.push_back(pos[y] - pos[x + 1] + 1);
    v.push_back(pos[x] - pos[y]);
    if (pos[x] < n - 1) v.push_back(n - 1 - pos[x]);
    res.push_back(v);
    vector<int> b;
    for (int i = pos[x] + 1; i < n; ++i) b.push_back(a[i]);
    for (int i = pos[y] + 1; i <= pos[x]; ++i) b.push_back(a[i]);
    for (int i = pos[x + 1]; i <= pos[y]; ++i) b.push_back(a[i]);
    for (int i = 0; i < pos[x + 1]; ++i) b.push_back(a[i]);
    a = b;
  }
  cout << res.size() << endl;
  for (auto v : res) {
    cout << v.size() << " ";
    for (auto e : v) cout << e << " ";
    cout << endl;
  }
  return 0;
}