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;
inline int gi() {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = (x << 3) + (x << 1) + (ch ^ 48);
    ch = getchar();
  }
  return (f == 1) ? x : -x;
}
inline int max(int a, int b) { return a > b ? a : b; }
inline int min(int a, int b) { return a < b ? a : b; }
const int maxn = 105;
int now[maxn], n;
int times = 0;
vector<vector<int> > ans;
inline void change(vector<int> k) {
  vector<int> temp;
  temp.clear();
  int cnt = 0;
  for (int i = 0; i <= k.size() - 1; ++i)
    if (k[i] != 0) cnt++;
  if (cnt == 1) return;
  ++times;
  int sta = 1;
  for (int i = 0; i <= k.size() - 1; ++i) {
    if (k[i] == 0) continue;
    int ed = sta + k[i] - 1;
    for (int j = ed; j >= sta; --j) temp.push_back(now[j]);
    sta = ed + 1;
  }
  reverse(temp.begin(), temp.end());
  for (int i = 0; i <= temp.size() - 1; ++i) now[i + 1] = temp[i];
  reverse(k.begin(), k.end());
  k.push_back(cnt);
  reverse(k.begin(), k.end());
  ans.push_back(k);
}
int main() {
  n = gi();
  for (int i = 1; i <= n; ++i) now[i] = gi();
  vector<int> vec;
  for (int i = 1; i <= n; ++i) {
    if (now[i] == 1) {
      vec.push_back(i - 1);
      vec.push_back(n - i + 1);
      break;
    }
  }
  change(vec);
  for (int i = 2; i <= n - 1; ++i) {
    if (i & 1) {
      int pos = 0;
      for (int j = 1; j <= n; ++j)
        if (now[j] == i) {
          pos = j;
          break;
        }
      vec.clear();
      vec.push_back(pos - 1);
      vec.push_back(n - pos - i + 2);
      for (int j = 1; j <= i - 1; ++j) vec.push_back(1);
      change(vec);
    } else {
      int pos = 0;
      for (int j = i; j <= n; ++j)
        if (now[j] == i) {
          pos = j;
          break;
        }
      vec.clear();
      for (int j = 1; j <= i - 1; ++j) vec.push_back(1);
      vec.push_back(pos - i + 1);
      vec.push_back(n - pos);
      change(vec);
    }
  }
  if (n & 1 && n != 1) {
    vec.clear();
    for (int i = 1; i <= n; ++i) vec.push_back(1);
    change(vec);
  }
  printf("%d\n", times);
  for (int i = 0; i <= times - 1; ++i) {
    for (int j = 0; j <= ans[i].size() - 1; ++j)
      if (ans[i][j] != 0) printf("%d ", ans[i][j]);
    printf("\n");
  }
  return 0;
}