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 N = 100000 + 10;
int n, c[N], idx = -1;
vector<vector<int> > op;
vector<int> tmp;
void print() {
  cout << op.size() << endl;
  for (int i = 0; i < op.size(); ++i) {
    printf("%d", (int)op[i].size());
    for (int j = 0; j < op[i].size(); ++j) printf(" %d", op[i][j]);
    puts("");
  }
}
void perform(vector<int> &f) {
  if (tmp.size() == 1) {
    tmp.clear();
    return;
  }
  int sum = 0;
  for (int i = 0; i < f.size(); ++i) sum += f[i];
  assert(sum == n);
  vector<vector<int> > chg;
  int idx = 0;
  for (int i = 0; i < f.size(); ++i) {
    chg.push_back(vector<int>());
    for (int tt = 0; tt < f[i]; ++tt) chg.back().push_back(c[idx++]);
  }
  idx = 0;
  reverse(chg.begin(), chg.end());
  for (auto g : chg)
    for (auto k : g) c[idx++] = k;
  op.push_back(f);
  f.clear();
}
void psh(int block) {
  if (block > 0) tmp.push_back(block);
}
void opr(int sorted, bool isLeft) {
  int target = n - sorted;
  for (int i = 0; i < sorted; ++i) {
    psh(1);
  }
  int idxTarget = -1;
  for (int i = 0; i < n; ++i)
    if (c[i] == target) idxTarget = i;
  psh(isLeft ? idxTarget - sorted + 1 : (n - idxTarget) - sorted);
  psh(isLeft ? n - idxTarget - 1 : idxTarget);
  if (!isLeft) reverse(tmp.begin(), tmp.end());
  perform(tmp);
}
void debugPrint() {
  for (int i = 0; i < n; ++i) {
    printf("%d ", c[i]);
  }
  puts("");
}
int main() {
  cin >> n;
  for (int i = 0; i < n; ++i) {
    scanf("%d", &c[i]);
    if (c[i] == n) idx = i;
  }
  psh(idx + (n % 2));
  psh(n - (idx + (n % 2)));
  perform(tmp);
  bool isLeft = !(n % 2);
  for (int sorted = 1; sorted < n; ++sorted) {
    opr(sorted, isLeft);
    isLeft = isLeft ^ 1;
  }
  print();
  return 0;
}