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 long long zero = 0;
const long long one = 1;
const long long INF = 9223372036854775807;
const int inINF = 2147483647;
const long long MOD = 1000000007;
const long long MOD2 = 998244353;
void Yes() { printf("Yes\n"); }
void No() { printf("No\n"); }
void YES() { printf("YES\n"); }
void NO() { printf("NO\n"); }
int main() {
  int n;
  cin >> n;
  vector<vector<int>> ans;
  vector<int> c(n);
  for (int i = 0; i < n; i++) {
    cin >> c[i];
    c[i]--;
  }
  int maxint, maxind, minint, minind;
  while (true) {
    bool check = true;
    for (int i = 0; i < n; i++) {
      if (c[i] != i) {
        check = false;
        break;
      }
    }
    if (check) break;
    bool ex, fin;
    fin = false;
    for (int i = 0; i < n - 1; i++) {
      ex = false;
      for (int j = 0; j < n; j++) {
        if (c[j] == i + 1) {
          ex = true;
        } else if (c[j] == i) {
          if (ex == false) {
            break;
          } else {
            maxint = i + 1;
            minint = i;
            fin = true;
            break;
          }
        }
      }
      if (fin) break;
    }
    int minlas, maxsta;
    for (int i = 0; i < n; i++) {
      if (c[i] == minint) {
        minlas = i;
      } else if (c[i] == maxint) {
        maxsta = i;
      }
    }
    minind = 0;
    for (int i = minlas; i > 0; i--) {
      if (c[i - 1] + 1 != c[i]) {
        minind = i;
        break;
      }
    }
    maxind = n - 1;
    for (int i = maxsta; i < n - 1; i++) {
      if (c[i] + 1 != c[i + 1]) {
        maxind = i;
        break;
      }
    }
    vector<int> d;
    if (maxsta == 0 && minlas == n - 1) {
      ans.push_back({minind - maxsta, minlas - minind + 1});
      for (int i = minind; i < n; i++) {
        d.push_back(c[i]);
      }
      for (int i = 0; i < minind; i++) {
        d.push_back(c[i]);
      }
    } else if (maxsta == 0) {
      ans.push_back({minind - maxsta, minlas - minind + 1, n - 1 - minlas});
      for (int i = minlas + 1; i < n; i++) {
        d.push_back(c[i]);
      }
      for (int i = minind; i < minlas + 1; i++) {
        d.push_back(c[i]);
      }
      for (int i = 0; i < minind; i++) {
        d.push_back(c[i]);
      }
    } else if (minlas == n - 1) {
      ans.push_back({maxsta, minind - maxsta, n - minind});
      for (int i = minind; i < n; i++) {
        d.push_back(c[i]);
      }
      for (int i = maxsta; i < minind; i++) {
        d.push_back(c[i]);
      }
      for (int i = 0; i < maxsta; i++) {
        d.push_back(c[i]);
      }
    } else {
      ans.push_back(
          {maxsta, minind - maxsta, minlas - minind + 1, n - minlas - 1});
      for (int i = minlas + 1; i < n; i++) {
        d.push_back(c[i]);
      }
      for (int i = minind; i < minlas + 1; i++) {
        d.push_back(c[i]);
      }
      for (int i = maxsta; i < minind; i++) {
        d.push_back(c[i]);
      }
      for (int i = 0; i < maxsta; i++) {
        d.push_back(c[i]);
      }
    }
    for (int i = 0; i < n; i++) {
      c[i] = d[i];
    }
  }
  int q = ans.size();
  printf("%d\n", q);
  for (int i = 0; i < q; i++) {
    int k = ans[i].size();
    printf("%d ", k);
    for (int j = 0; j < ans[i].size() - 1; j++) {
      printf("%d ", ans[i][j]);
    }
    printf("%d\n", ans[i][ans[i].size() - 1]);
  }
}