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 inf = 132913423200339;
const long long mod = 998244353;
const int N = 1e6 + 5;
const long long M = 1e9;
const double eps = 1e-7;
using namespace std;
inline long long ksc(long long x, long long y, long long p) {
  long long res = 0;
  while (y) {
    if (y & 1) res = (res + x) % p;
    x = (x << 1) % p;
    y >>= 1;
  }
  return res;
}
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
long long lcm(long long a, long long b) { return a * b / gcd(a, b); }
long long qp(long long a, long long b, long long p) {
  long long ans = 1;
  while (b) {
    if (b & 1) {
      ans = (ans * a) % p;
      --b;
    }
    a = (a * a) % p;
    b >>= 1;
  }
  return ans % p;
}
long long Inv(long long x) { return qp(x, mod - 2, mod); }
long long C(long long n, long long m) {
  if (m > n) return 0;
  long long ans = 1;
  for (int i = 1; i <= m; ++i) ans = ans * Inv(i) % mod * (n - i + 1) % mod;
  return ans % mod;
}
long long A(long long n, long long m, long long mod) {
  long long sum = 1;
  for (int i = n; i >= n - m + 1; i--) sum = (sum * i) % mod;
  return sum % mod;
}
long long lowbit(long long x) { return x & (-x); }
int a[100], b[100];
vector<int> v[10005];
vector<int> ans[10005];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  int id = 0;
  while (true) {
    id++;
    int cnt = 1, f, p;
    int flag = 0;
    for (int i = 2; i <= n; i++) {
      if (a[i - 1] + 1 != a[i]) flag = 10;
    }
    if (!flag) break;
    for (int i = 1; i <= n; i++) {
      if (a[i] != i) {
        f = i;
        break;
      }
    }
    if (f != 1) ans[id].push_back(f - 1);
    for (int i = f + 1; i <= n; i++) {
      if (a[i - 1] + 1 == a[i])
        cnt++;
      else {
        p = a[f];
        f = i;
        break;
      }
    }
    ans[id].push_back(cnt);
    cnt = 1;
    for (int i = f; i <= n; i++) {
      if (a[i] == p - 1) {
        ans[id].push_back(cnt);
        f = i;
        break;
      }
      cnt++;
    }
    if (f != n) ans[id].push_back(n - f);
    int pos = n, w = 1;
    for (int i = ans[id].size() - 1; i >= 0; i--) {
      for (int j = pos - ans[id][i] + 1; j <= pos; j++) {
        b[w++] = a[j];
      }
      pos -= ans[id][i];
    }
    for (int i = 1; i <= n; i++) a[i] = b[i];
  }
  printf("%d\n", id - 1);
  for (int i = 1; i <= id - 1; i++) {
    printf("%d ", ans[i].size());
    for (int j = 0; j < ans[i].size(); j++) printf("%d ", ans[i][j]);
    puts("");
  }
  return 0;
}