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 main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int tt = 1;
  mt19937 rng(58);
  while (tt--) {
    int N;
    cin >> N;
    vector<int> P(N);
    for (int &p : P) cin >> p, --p;
    vector<vector<int>> ops;
    auto make = [&](vector<int> d) {
      if (d.size() < 2) return;
      ops.emplace_back(d);
      vector<vector<int>> parts;
      int cur = 0;
      for (int v : d) {
        parts.emplace_back(vector<int>(P.begin() + cur, P.begin() + cur + v));
        cur += v;
      }
      assert(cur == N);
      reverse(parts.begin(), parts.end());
      int i = 0;
      for (auto vv : parts)
        for (int v : vv) P[i++] = v;
      assert(i == N);
    };
    for (int i = 1; i < N; ++i) {
      int cur = find(P.begin(), P.end(), 0) - P.begin();
      bool rev = false;
      if (cur > 0 && i > 1 && P[cur - 1] == 1) rev = true;
      if (rev) {
        int nxt = find(P.begin(), P.end(), i) - P.begin();
        if (nxt < cur) {
          vector<int> d;
          if (nxt) d.emplace_back(nxt);
          d.emplace_back(cur - (i - 1) - nxt);
          for (int z = 0; z < i; ++z) {
            d.emplace_back(1);
          }
          if (cur < N - 1) d.emplace_back(N - 1 - cur);
          make(d);
        } else {
          vector<int> d;
          if (cur - (i - 1) > 0) d.emplace_back(cur - (i - 1));
          d.emplace_back(i);
          d.emplace_back(nxt - cur);
          if (nxt < N - 1) d.emplace_back(N - 1 - nxt);
          make(d);
        }
      } else {
        int nxt = find(P.begin(), P.end(), i) - P.begin();
        if (nxt < cur) {
          vector<int> d;
          if (nxt > 0) d.emplace_back(nxt);
          d.emplace_back(1);
          d.emplace_back(cur + i - 1 - nxt);
          if (cur + i - 1 < N - 1) d.emplace_back(N - 1 - (cur + i - 1));
          make(d);
        } else {
          vector<int> d;
          if (cur > 0) d.emplace_back(cur);
          for (int z = 0; z < i; ++z) d.emplace_back(1);
          d.emplace_back(nxt - (cur + i - 1));
          if (nxt < N - 1) d.emplace_back(N - 1 - nxt);
          make(d);
        }
      }
    }
    if (N > 1 && P[0] == N - 1) {
      vector<int> d(N, 1);
      make(d);
    }
    assert(is_sorted(P.begin(), P.end()));
    cout << ops.size() << '\n';
    for (auto op : ops) {
      cout << op.size() << ' ';
      for (int v : op) cout << v << ' ';
      cout << '\n';
    }
  }
}