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 = 55;
const int inf = 2147483647;
const double pi = acos(-1.0);
int read() {
  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 * 10 + ch - '0', ch = getchar();
  return x * f;
}
int n, a[Maxn], tmp[Maxn], lb, pos[Maxn];
vector<int> ans[Maxn];
int cnt = 0;
struct T {
  int c, v, l, r;
} b[Maxn];
int main() {
  n = read();
  for (int i = 1; i <= n; i++) a[i] = read();
  while (1) {
    int cur = 1;
    b[lb = 1].l = b[1].r = b[1].c = 1;
    b[1].v = a[1];
    for (int i = 2; i <= n; i++) {
      if (a[i] == a[i - 1] + 1)
        b[lb].c++, b[lb].r = i;
      else
        b[++lb].c = 1, b[lb].l = b[lb].r = i, b[lb].v = a[i];
    }
    if (lb == 1) break;
    cnt++;
    for (int i = 1; i <= lb; i++) {
      int p, mn = inf;
      for (int j = 1; j <= lb; j++)
        if (b[j].v >= i && b[j].v <= mn) mn = b[j].v, p = j;
      b[p].v = i;
      pos[i] = p;
    }
    for (int i = 2; i <= lb; i++)
      if (pos[i] < pos[i - 1]) {
        if (b[pos[i]].l != 1) ans[cnt].push_back(b[pos[i]].l - 1);
        ans[cnt].push_back(b[pos[i]].r - b[pos[i]].l + 1);
        ans[cnt].push_back(b[pos[i - 1]].r - b[pos[i]].r);
        if (b[pos[i - 1]].r != n) ans[cnt].push_back(n - b[pos[i - 1]].r);
        break;
      }
    int t = 0;
    for (int i = 0; i < ans[cnt].size(); i++) {
      for (int j = t + 1; j <= t + ans[cnt][i]; j++)
        tmp[n - t - ans[cnt][i] + j - t] = a[j];
      t += ans[cnt][i];
    }
    for (int i = 1; i <= n; i++) a[i] = tmp[i];
  }
  printf("%d\n", cnt);
  for (int i = 1; i <= cnt; i++) {
    printf("%d ", ans[i].size());
    for (int j = 0; j < ans[i].size(); j++) printf("%d ", ans[i][j]);
    puts("");
  }
}