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 MAXM = 1000001;
int main() {
  int n;
  scanf("%d", &n);
  vector<int> c(n);
  vector<vector<int> > res;
  for (int i = 0; i < n; i++) scanf("%d", &c[i]);
  while (!is_sorted(c.begin(), c.end())) {
    if (is_sorted(c.rbegin(), c.rend())) {
      res.push_back(vector<int>(n, 1));
      reverse(c.begin(), c.end());
      continue;
    }
    int pos = find(c.begin(), c.end(), 1) - c.begin();
    int npos = find(c.begin(), c.end(), 2) - c.begin();
    int lb, rb;
    int needReverse;
    int target;
    if (abs(pos - npos) == 1) {
      int sign = pos < npos ? 1 : -1;
      while (npos >= 0 && npos < c.size() && c[npos] == abs(npos - pos) + 1)
        npos += sign;
      npos -= sign;
      needReverse = npos < pos;
      lb = min(pos, npos);
      rb = max(npos, pos);
      target = find(c.begin(), c.end(), c[npos] + 1) - c.begin();
    } else {
      lb = pos;
      rb = pos;
      needReverse = false;
      target = npos;
    }
    pos = lb;
    npos = rb;
    assert(pos != target);
    vector<int> cur;
    if (pos < target && !needReverse) {
      cur.push_back(pos);
      for (int i = 0; i < abs(pos - npos) + 1; i++) cur.push_back(1);
      cur.push_back(target - npos);
      cur.push_back(n - target - 1);
    } else if (pos < target && needReverse) {
      cur.push_back(pos);
      cur.push_back(abs(pos - npos) + 1);
      cur.push_back(target - npos);
      cur.push_back(n - target - 1);
    } else if (pos > target && !needReverse) {
      cur.push_back(target);
      cur.push_back(1);
      cur.push_back(max(pos, npos) - target);
      cur.push_back(n - max(pos, npos) - 1);
    } else {
      cur.push_back(target);
      cur.push_back(min(pos, npos) - target);
      for (int i = 0; i < abs(pos - npos) + 1; i++) cur.push_back(1);
      cur.push_back(n - max(pos, npos) - 1);
    }
    cur.erase(remove_if(cur.begin(), cur.end(), [](int x) { return x == 0; }),
              cur.end());
    res.push_back(cur);
    vector<int> nc;
    int top = 0;
    for (int sz : cur) {
      nc.insert(nc.begin(), c.begin() + top, c.begin() + top + sz);
      top += sz;
    }
    c = nc;
  }
  printf("%d\n", (int)res.size());
  for (const vector<int>& cur : res) {
    printf("%d", (int)cur.size());
    for (int x : cur) printf(" %d", x);
    puts("");
  }
  return 0;
}