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 INF = 1000000007;
const long long MOD = 1000000007;
vector<vector<pair<int, int>>> ans;
vector<int> c(52);
void solve_test() {
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> c[i];
  }
  if (n == 1) {
    cout << 0;
    return;
  }
  for (int num = 0; num < n - 1; num++) {
    bool sorted = true;
    for (int i = 1; i < n; i++) {
      sorted &= c[i] > c[i - 1];
    }
    if (sorted) break;
    vector<pair<int, int>> v;
    int target = n - num - 1;
    bool seen_target = false, seen_block = false;
    bool seen[n];
    memset(seen, false, sizeof(seen));
    bool need = true;
    for (int i = 0; i < n; i++) {
      if (c[i] == target + 1) {
        if (i < n && c[i + 1] == target) need = false;
        if (i && c[i - 1] == target) need = false;
      }
    }
    if (!need) continue;
    int next_pos = 0;
    for (int i = 0; i < n; i++) {
      if (c[i] == target) {
        if (seen_block) {
          v.push_back({next_pos, i - next_pos + 1});
          for (int j = next_pos; j <= i; j++) {
            seen[j] = true;
          }
        } else {
          int j = i;
          for (; j < n && (c[j] != n && c[j] != n - num); j++) {
            seen[j] = true;
          }
          v.push_back({i, j - i});
        }
        seen_target = true;
      }
      if (!seen_block && (c[i] == n || c[i] == n - num)) {
        if ((c[i] == n) ^ seen_target) {
          int j = i;
          for (; j < n && c[j] >= n - num; j++) {
            seen[j] = true;
            v.push_back({j, 1});
          }
          next_pos = j;
        } else {
          int j = i;
          for (; j < n && c[j] <= n && c[j] >= n - num; j++) {
            seen[j] = true;
          }
          next_pos = j;
          v.push_back({i, next_pos - i});
        }
        seen_block = true;
      }
    }
    for (int i = 0; i < n; i++) {
      if (!seen[i]) {
        v.push_back({i, 1});
      }
    }
    sort(v.begin(), v.end());
    vector<int> c_(n);
    for (int i = 0; i < n; i++) {
      c_[i] = c[i];
    }
    int j = 0;
    for (int i = (int)v.size() - 1; i >= 0; i--) {
      for (int st = v[i].first; st < v[i].first + v[i].second; st++) {
        c[j++] = c_[st];
      }
    }
    ans.push_back(v);
  }
  if (c[0] == n) {
    vector<pair<int, int>> v;
    for (int i = 0; i < n; i++) {
      v.push_back({i, 1});
    }
    ans.push_back(v);
  }
  cout << ans.size() << "\n";
  for (vector<pair<int, int>> v : ans) {
    cout << (int)v.size() << " ";
    for (pair<int, int> x : v) {
      cout << x.second << " ";
    }
    cout << "\n";
  }
}
signed main() {
  int tests;
  tests = 1;
  while (tests--) {
    solve_test();
  }
  return 0;
}