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 = 52;
int n;
vector<pair<int, pair<int, int> > > a;
struct Answer {
  vector<pair<int, int> > a;
} ans[Maxn * Maxn + 5];
int len;
pair<int, int> Merge(pair<int, int> l, pair<int, int> r) {
  return make_pair(min(l.first, r.first), max(l.second, r.second));
}
pair<int, int> Reverse(pair<int, int> x) {
  return make_pair(n - x.second + 1, n - x.first + 1);
}
void work(vector<pair<int, pair<int, int> > > a) {
  if (a.size() == 1) {
    return;
  }
  int pos_1, pos_2;
  for (int i = 1; i < (int)a.size(); i++) {
    pos_1 = pos_2 = -1;
    for (int j = 0; j < (int)a.size(); j++) {
      if (a[j].first == i) {
        pos_1 = j;
      }
      if (a[j].first == i + 1) {
        pos_2 = j;
      }
    }
    if (pos_2 == pos_1 + 1) {
      vector<pair<int, pair<int, int> > > nxt;
      for (int j = 0; j < pos_1; j++) {
        if (a[j].first <= i) {
          nxt.push_back(a[j]);
        } else {
          nxt.push_back(make_pair(a[j].first - 1, a[j].second));
        }
      }
      nxt.push_back(
          make_pair(a[pos_1].first, Merge(a[pos_1].second, a[pos_2].second)));
      for (int j = pos_2 + 1; j < (int)a.size(); j++) {
        if (a[j].first <= i) {
          nxt.push_back(a[j]);
        } else {
          nxt.push_back(make_pair(a[j].first - 1, a[j].second));
        }
      }
      work(nxt);
      return;
    }
    if (pos_1 < pos_2) {
      continue;
    }
    len++;
    if (pos_2 != 0) {
      ans[len].a.push_back(make_pair(1, a[pos_2].second.first - 1));
    }
    ans[len].a.push_back(
        make_pair(a[pos_2].second.first, a[pos_1].second.first - 1));
    ans[len].a.push_back(a[pos_1].second);
    if (pos_1 != (int)a.size() - 1) {
      ans[len].a.push_back(make_pair(a[pos_1].second.second + 1, n));
    }
    vector<pair<int, pair<int, int> > > nxt;
    if (pos_1 != (int)a.size() - 1) {
      for (int j = pos_1 + 1; j < (int)a.size(); j++) {
        if (a[j].first <= i) {
          nxt.push_back(make_pair(
              a[j].first,
              make_pair(a[j].second.first - a[pos_1].second.second,
                        a[j].second.second - a[pos_1].second.second)));
        } else {
          nxt.push_back(make_pair(
              a[j].first - 1,
              make_pair(a[j].second.first - a[pos_1].second.second,
                        a[j].second.second - a[pos_1].second.second)));
        }
      }
    }
    nxt.push_back(
        make_pair(a[pos_1].first,
                  make_pair(n - a[pos_1].second.second + 1,
                            a[pos_2].second.second - a[pos_2].second.first + 1 +
                                n - a[pos_1].second.first + 1)));
    for (int j = pos_2 + 1; j < pos_1; j++) {
      if (a[j].first <= i) {
        nxt.push_back(make_pair(
            a[j].first,
            make_pair(a[j].second.first - a[pos_2].second.second + n -
                          a[pos_1].second.first + 1 + a[pos_2].second.second -
                          a[pos_2].second.first + 1,
                      a[j].second.second - a[pos_2].second.second + n -
                          a[pos_1].second.first + 1 + a[pos_2].second.second -
                          a[pos_2].second.first + 1)));
      } else {
        nxt.push_back(make_pair(
            a[j].first - 1,
            make_pair(a[j].second.first - a[pos_2].second.second + n -
                          a[pos_1].second.first + 1 + a[pos_2].second.second -
                          a[pos_2].second.first + 1,
                      a[j].second.second - a[pos_2].second.second + n -
                          a[pos_1].second.first + 1 + a[pos_2].second.second -
                          a[pos_2].second.first + 1)));
      }
    }
    if (pos_2 != 0) {
      int len = n - a[pos_2].second.first + 1;
      for (int j = 0; j < pos_2; j++) {
        if (a[j].first <= i) {
          nxt.push_back(make_pair(
              a[j].first,
              make_pair(a[j].second.first + len, a[j].second.second + len)));
        } else {
          nxt.push_back(make_pair(
              a[j].first - 1,
              make_pair(a[j].second.first + len, a[j].second.second + len)));
        }
      }
    }
    work(nxt);
    return;
  }
}
int main() {
  scanf("%d", &n);
  int x;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &x);
    a.push_back(make_pair(x, make_pair(i, i)));
  }
  work(a);
  printf("%d\n", len);
  for (int i = 1; i <= len; i++) {
    printf("%d ", ans[i].a.size());
    for (int j = 0; j < (int)ans[i].a.size(); j++) {
      printf("%d ", ans[i].a[j].second - ans[i].a[j].first + 1);
    }
    puts("");
  }
  return 0;
}