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 maxn = 2e5 + 10;
const int mod = 1e9 + 7;
int n;
struct block {
  vector<int> v;
  int id;
};
vector<block> b;
int a[maxn];
void get_blocks() {
  b.clear();
  block cur;
  int j = 1;
  while (j <= n) {
    cur.v.clear();
    while (1) {
      cur.v.push_back(a[j]);
      if (j == n || a[j + 1] != a[j] + 1) {
        ++j;
        break;
      }
      ++j;
    }
    b.push_back(cur);
  }
  vector<int> bv;
  for (auto& w : b) bv.push_back(w.v[0]);
  sort(bv.begin(), bv.end());
  for (auto& w : b) {
    int u = lower_bound(bv.begin(), bv.end(), w.v[0]) - bv.begin() + 1;
    w.id = u;
  }
}
vector<vector<int>> tv;
void tr() {
  vector<vector<block>> seq;
  int la = b.back().id;
  int en = b.size() - 1;
  vector<block> last;
  last.push_back(b.back());
  vector<int> e;
  vector<block> mid;
  if (la == b.size()) {
    la = b[b.size() - 2].id;
    vector<block> q;
    q.push_back(b[b.size() - 2]);
    --en;
    for (int i = 0; i < b.size(); ++i) {
      if (b[i].id == la + 1) {
        for (int j = i; j < en; ++j) mid.push_back(b[j]);
        vector<block> fr;
        for (int j = 0; j < i; ++j) fr.push_back(b[j]);
        if (fr.size()) seq.push_back(fr);
        seq.push_back(mid);
        seq.push_back(q);
        seq.push_back(last);
        break;
      }
    }
    goto N;
  }
  for (int i = 0; i < b.size(); ++i) {
    if (b[i].id == la + 1) {
      for (int j = i; j < en; ++j) mid.push_back(b[j]);
      vector<block> fr;
      for (int j = 0; j < i; ++j) fr.push_back(b[j]);
      if (fr.size()) seq.push_back(fr);
      seq.push_back(mid);
      seq.push_back(last);
      break;
    }
  }
N:;
  e.push_back(seq.size());
  for (auto& w : seq) {
    int cnt = 0;
    for (auto& bb : w) cnt += bb.v.size();
    e.push_back(cnt);
  }
  tv.push_back(e);
  reverse(seq.begin(), seq.end());
  int j = 1;
  for (auto& w : seq) {
    for (auto& ee : w) {
      for (auto& ii : ee.v) a[j++] = ii;
    }
  }
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; ++i) cin >> a[i];
  while (1) {
    get_blocks();
    if (b.size() == 1) break;
    tr();
  }
  cout << tv.size() << "\n";
  for (auto& w : tv) {
    for (int i = 0; i < w.size(); ++i) {
      cout << w[i];
      if (i < w.size() - 1)
        cout << " ";
      else
        cout << "\n";
    }
  }
}