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];
vector<vector<int> > op;
bool isSorted() {
  for (int i = 0; i < n; ++i) {
    if (c[i] != i + 1) return false;
  }
  return true;
}
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) {
  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 debugPrint() {
  for (int i = 0; i < n; ++i) {
    printf("%d ", c[i]);
  }
  puts("");
}
void rght(int sorted) {
  int target = n - sorted;
  vector<int> tmp;
  int idx = 0;
  while (idx < sorted) {
    tmp.push_back(1);
    idx++;
  }
  int cnt = 0;
  while (c[idx] != target) {
    cnt++;
    idx++;
  }
  cnt++;
  idx++;
  tmp.push_back(cnt);
  cnt = n - idx;
  if (cnt > 0) tmp.push_back(cnt);
  perform(tmp);
}
void lft(int sorted) {
  int target = n - sorted;
  vector<int> tmp;
  int idx = n - 1;
  while (idx >= n - sorted) {
    tmp.push_back(1);
    idx--;
  }
  int cnt = 0;
  while (c[idx] != target) {
    cnt++;
    idx--;
  }
  cnt++;
  idx--;
  tmp.push_back(cnt);
  cnt = idx + 1;
  if (cnt > 0) tmp.push_back(cnt);
  reverse(tmp.begin(), tmp.end());
  perform(tmp);
}
int main() {
  cin >> n;
  for (int i = 0; i < n; ++i) {
    scanf("%d", &c[i]);
  }
  if (isSorted()) {
    puts("0");
    return 0;
  }
  bool left = true;
  vector<int> tmp;
  if (n % 2 == 0) {
    if (c[0] != n) {
      for (int i = 0; i < n; ++i) {
        if (c[i] == n) {
          tmp.push_back(i);
          tmp.push_back(n - i);
        }
      }
      perform(tmp);
    }
    left = true;
  } else {
    if (c[n - 1] != n) {
      for (int i = 0; i < n; ++i) {
        if (c[i] == n) {
          tmp.push_back(i + 1);
          tmp.push_back(n - i - 1);
        }
      }
      perform(tmp);
    }
    left = false;
  }
  for (int sorted = 1; sorted < n; ++sorted) {
    if (left) {
      rght(sorted);
    } else {
      lft(sorted);
    }
    left = !left;
  }
  assert(op.size() <= n);
  print();
  return 0;
}