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>
#pragma GCC optimize("-O3")
using namespace std;
long long MOD = 998244353;
long long mpow(long long a, long long b) {
  if (b >= (MOD - 1)) b %= (MOD - 1);
  if (b == 0) return 1;
  long long t1 = mpow(a, b / 2);
  t1 *= t1;
  t1 %= MOD;
  if (b % 2) t1 *= a;
  t1 %= MOD;
  return t1;
}
long long mpow(long long a, long long b, long long p) {
  if (b == 0) return 1;
  long long t1 = mpow(a, b / 2, p);
  t1 *= t1;
  t1 %= p;
  if (b % 2) t1 *= a;
  t1 %= p;
  return t1;
}
long long modinverse(long long a, long long m) {
  long long m0 = m;
  long long y = 0, x = 1;
  if (m == 1) return 0;
  while (a > 1) {
    long long q = a / m;
    long long t = m;
    m = a % m, a = t;
    t = y;
    y = x - q * y;
    x = t;
  }
  if (x < 0) x += m0;
  return x;
}
long long range(long long l, long long r) { return l + rand() % (r - l + 1); }
pair<long long, long long> hashFn(vector<long long> vc) {
  long long p1 = 31, p2 = 97;
  long long m1 = 1e9 + 9, m2 = 998244353;
  long long hsv1 = 0, hsv2 = 0, p_pow1 = 1, p_pow2 = 1;
  for (long long i = 0; i < vc.size(); i++) {
    hsv1 = ((hsv1 + vc[i] + 1) * p_pow1) % m1;
    p_pow1 = (p_pow1 * p1) % m1;
    hsv2 = ((hsv2 + vc[i] + 1) * p_pow2) % m2;
    p_pow2 = (p_pow2 * p2) % m2;
  }
  return make_pair(hsv1, hsv2);
}
void solve() {
  long long n;
  cin >> n;
  vector<long long> c(n);
  vector<long long> pos(n);
  for (long long i = 0; i < n; i++) {
    cin >> c[i];
    c[i]--;
    pos[c[i]] = i;
  }
  long long ans = 0;
  vector<vector<long long> > seq;
  for (long long l = 0; l < n; l++) {
    vector<long long> blk(n, -1);
    for (long long i = 0; i < n; i++) {
      long long st = pos[i];
      long long j = 0;
      while (st + j < n && i + j == c[st + j]) j++;
      blk[i] = i + j;
      i = i + j - 1;
    }
    for (long long i = 0; i < n - 1; i++) {
      if (blk[i] != -1 && pos[blk[i]] < pos[i]) {
        ans++;
        vector<long long> elms;
        long long s = 0;
        if (pos[blk[i]] != 0) {
          elms.push_back(pos[blk[i]]);
          s += pos[blk[i]];
        }
        elms.push_back(pos[i] - pos[blk[i]]);
        s += pos[i] - pos[blk[i]];
        elms.push_back(blk[i] - i);
        s += blk[i] - i;
        if (s < n) elms.push_back(n - s);
        seq.push_back(elms);
        vector<long long> nc(n);
        long long curr = n - 1, curr2 = 0;
        for (long long j = 0; j < elms.size(); j++) {
          for (long long k = 0; k < elms[j]; k++) {
            nc[curr - elms[j] + k + 1] = c[k + curr2];
          }
          curr = curr - elms[j];
          curr2 += elms[j];
        }
        c = nc;
        for (long long l = 0; l < n; l++) {
          pos[c[l]] = l;
        }
        break;
      }
    }
  }
  cout << ans << "\n";
  for (long long i = 0; i < ans; i++) {
    cout << seq[i].size() << " ";
    for (long long j = 0; j < seq[i].size(); j++) {
      cout << seq[i][j] << " ";
    }
    cout << "\n";
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t = 1;
  for (long long i = 0; i < t; i++) {
    solve();
  }
  return 0;
}