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 maxn = 1e6 + 50;
const int mod = 1e9 + 7;
int n, m;
int a[maxn], b[maxn];
vector<int> vec, vec2[maxn];
int stk[maxn];
int main(int argc, char const *argv[]) {
  int tot2 = 0;
  int pre = 0;
  scanf("%d", &n);
  for (int i = (1); i <= (n); ++i) {
    scanf("%d", &a[i]);
    if (a[i] == 1) pre = i;
  }
  int p = 2;
  int tot = n;
  int fg = 1;
  while (tot--) {
    if (p >= n + 1) {
      if (!fg) {
        tot2++;
        for (int i = (1); i <= (n); ++i) {
          vec2[tot2].push_back(1);
        }
      }
      break;
    }
    vec.clear();
    int st = 0;
    for (int i = (1); i <= (n); ++i) {
      if (a[i] == p) {
        st = i;
        break;
      }
    }
    int cnt = 0;
    if (fg) {
      if (st > pre) {
        vec.push_back(pre - 1);
        for (int i = (1); i <= (p - 1); ++i) {
          vec.push_back(1);
        }
        vec.push_back(st - (pre + p - 2));
        vec.push_back(n - st);
        int len = vec.size();
        int len2 = len;
        int le = 1;
        for (int i = (0); i <= (len - 1); ++i) {
          for (int j = (le + vec[i] - 1); j >= (le); --j) {
            stk[++cnt] = a[j];
          }
          le += vec[i];
        }
        cnt = 0;
        for (int i = (n); i >= (1); --i) {
          a[++cnt] = stk[i];
          if (a[cnt] == p) pre = cnt;
        }
        fg = 0;
      } else {
        vec.push_back(st - 1);
        vec.push_back(pre - st);
        vec.push_back(p - 1);
        vec.push_back(n - (pre + p - 2));
        int len = vec.size();
        int len2 = len;
        int le = 1;
        for (int i = (0); i <= (len - 1); ++i) {
          for (int j = (le + vec[i] - 1); j >= (le); --j) {
            stk[++cnt] = a[j];
          }
          le += vec[i];
        }
        cnt = 0;
        for (int i = (n); i >= (1); --i) {
          a[++cnt] = stk[i];
          if (a[cnt] == 1) pre = cnt;
        }
      }
    } else {
      if (st > pre) {
        vec.push_back(pre - 1);
        vec.push_back(p - 1);
        vec.push_back(st - (pre + p - 2));
        vec.push_back(n - st);
        int len = vec.size();
        int len2 = len;
        int le = 1;
        for (int i = (0); i <= (len - 1); ++i) {
          for (int j = (le + vec[i] - 1); j >= (le); --j) {
            stk[++cnt] = a[j];
          }
          le += vec[i];
        }
        cnt = 0;
        for (int i = (n); i >= (1); --i) {
          a[++cnt] = stk[i];
          if (a[cnt] == p) pre = cnt;
        }
      } else {
        vec.push_back(st - 1);
        vec.push_back(pre - st);
        for (int i = (1); i <= (p - 1); ++i) {
          vec.push_back(1);
        }
        vec.push_back(n - (pre + p - 2));
        int len = vec.size();
        int len2 = len;
        int le = 1;
        for (int i = (0); i <= (len - 1); ++i) {
          for (int j = (le + vec[i] - 1); j >= (le); --j) {
            stk[++cnt] = a[j];
          }
          le += vec[i];
        }
        cnt = 0;
        for (int i = (n); i >= (1); --i) {
          a[++cnt] = stk[i];
          if (a[cnt] == 1) pre = cnt;
        }
        fg = 1;
      }
    }
    tot2++;
    int len = vec.size();
    for (int i = (0); i <= (len - 1); ++i) {
      if (vec[i]) {
        vec2[tot2].push_back(vec[i]);
      }
    }
    p++;
  }
  printf("%d\n", tot2);
  for (int i = (1); i <= (tot2); ++i) {
    int len = vec2[i].size();
    printf("%d ", len);
    for (int j = (0); j <= (len - 1); ++j) {
      printf("%d ", vec2[i][j]);
    }
    printf("\n");
  }
  return 0;
}