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;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
inline long long rand(long long x, long long y) {
  return rng() % (y + 1 - x) + x;
}
string inline to_string(char c) {
  string second(1, c);
  return second;
}
template <typename T>
inline T gcd(T a, T b) {
  return a == 0 ? llabs(b) : gcd(b % a, a);
}
using ll = long long;
using ld = long double;
using pi = pair<ll, ll>;
using spi = pair<ll, pi>;
using dpi = pair<pi, pi>;
vector<ll> v;
bool inv;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  ll n;
  cin >> n;
  v.resize(n);
  for (ll i = 0; i <= ll(n - 1); ++i) {
    cin >> v[i];
  }
  vector<vector<ll>> Q;
  for (ll i = 2; i <= ll(n); ++i) {
    ll pos1 = find((v).begin(), (v).end(), i - 1) - v.begin();
    ll pos2 = find((v).begin(), (v).end(), i) - v.begin();
    if (pos1 > pos2) {
      vector<ll> q;
      if (pos2) q.emplace_back(pos2);
      q.emplace_back(pos1 - pos2);
      if (inv)
        for (ll j = 1; j <= ll(i - 1); ++j) q.emplace_back(1);
      else
        q.emplace_back(i - 1);
      inv = 0;
      q.emplace_back(ll(v.size()) - pos1 - 1);
      Q.emplace_back(q);
      vector<ll> tmp;
      for (ll i = pos1 + 1; i <= ll(ll(v.size()) - 1); ++i) {
        tmp.emplace_back(v[i]);
      }
      tmp.emplace_back(i);
      for (ll i = pos2 + 1; i <= ll(pos1 - 1); ++i) {
        tmp.emplace_back(v[i]);
      }
      for (ll i = 0; i <= ll(pos2 - 1); ++i) {
        tmp.emplace_back(v[i]);
      }
      v = tmp;
    } else {
      vector<ll> q;
      if (pos1) q.emplace_back(pos1);
      if (inv)
        q.emplace_back(i - 1);
      else
        for (ll j = 1; j <= ll(i - 1); ++j) q.emplace_back(1);
      inv = 1;
      q.emplace_back(pos2 - pos1);
      q.emplace_back(ll(v.size()) - pos2 - 1);
      Q.emplace_back(q);
      vector<ll> tmp;
      for (ll i = pos2 + 1; i <= ll(ll(v.size()) - 1); ++i) {
        tmp.emplace_back(v[i]);
      }
      for (ll i = pos1 + 1; i <= ll(pos2); ++i) {
        tmp.emplace_back(v[i]);
      }
      for (ll i = 0; i <= ll(pos1 - 1); ++i) {
        tmp.emplace_back(v[i]);
      }
      v = tmp;
    }
  }
  if (inv) {
    Q.emplace_back(vector<ll>(n, 1));
  }
  cout << Q.size() << '\n';
  for (auto i : Q) {
    ll co = 0;
    for (auto j : i) {
      assert(j >= 0);
      co += j > 0;
    }
    cout << co << ' ';
    for (auto j : i)
      if (j) cout << j << ' ';
    cout << '\n';
  }
}