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 I1 = 1e9;
const long long I2 = 1e18;
const int32_t M1 = 1e9 + 7;
const int32_t M2 = 998244353;
template <typename T, typename T1>
T maxn(T &a, T1 b) {
  if (b > a) a = b;
  return a;
}
template <typename T, typename T1>
T minn(T &a, T1 b) {
  if (b < a) a = b;
  return a;
}
vector<vector<long long int> > v;
void solve() {
  int n;
  cin >> n;
  vector<long long int> a(n), b(n);
  for (long long int i = 0; i < n; i++) b[i] = i + 1;
  for (long long int i = 0; i < n; i++) cin >> a[i];
  while (1) {
    if (a == b) break;
    int mm = a[n - 1];
    bool cc = false;
    int cnt = 1, cnt1 = 0;
    vector<long long int> vv;
    for (long long int i = n - 1 - 1; i >= 0; i--) {
      if (a[i + 1] - a[i] == 1 && !cc) {
        cnt++;
        continue;
      } else {
        cnt1 += cnt;
        vv.push_back(cnt);
        cc = true;
        if (mm != n) break;
        mm = a[i];
        cnt = 1;
        i--;
        while (a[i + 1] - a[i] == 1 && i >= 0) {
          cnt++;
          i--;
        }
        vv.push_back(cnt);
        cnt1 += cnt;
        break;
      }
    }
    int ind = -1;
    for (long long int i = 0; i < n; i++) {
      if (a[i] == mm + 1) {
        ind = i;
        break;
      }
    }
    vv.push_back(n - cnt1 - ind);
    cnt1 += (n - cnt1 - ind);
    if (cnt1 != n) vv.push_back(n - cnt1);
    vector<long long int> temp;
    int p = n;
    for (auto u : vv) {
      for (int ii = p - u; ii < p; ii++) {
        temp.push_back(a[ii]);
      }
      p -= u;
    }
    reverse(vv.begin(), vv.end());
    v.push_back(vv);
    a.clear();
    a = temp;
  }
  cout << (long long int)v.size() << '\n';
  for (auto u : v) {
    cout << (long long int)u.size() << " ";
    for (auto uu : u) cout << uu << " ";
    cout << '\n';
  }
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int testcase = 1;
  while (testcase--) solve();
  return 0;
}