Problem: Snuke has a sequence p, which is a permutation of (0,1,2, ...,N-1). The i-th element (0-indexed) in p is p_i.

He can perform N-1 kinds of operations labeled 1,2,...,N-1 any number of times in any order. When the operation labeled k is executed, the procedure represented by the following code will be performed:


for(int i=k;i<N;i++)
swap(p[i],p[i-k]);


He would like to sort p in increasing order using between 0 and 10^{5} operations (inclusive). Show one such sequence of operations. It can be proved that there always exists such a sequence of operations under the constraints in this problem.

Constraints

* 2 \leq N \leq 200
* 0 \leq p_i \leq N-1
* p is a permutation of (0,1,2,...,N-1).

Input

Input is given from Standard Input in the following format:


N
p_0 p_1 ... p_{N-1}


Output

Let m be the number of operations in your solution. In the first line, print m. In the i-th of the following m lines, print the label of the i-th executed operation. The solution will be accepted if m is at most 10^5 and p is in increasing order after the execution of the m operations.

Examples

Input

5
4 2 0 1 3


Output

4
2
3
1
2


Input

9
1 0 4 3 5 6 2 8 7


Output

11
3
6
1
3
5
2
4
7
8
6
3

Solution (in CPP):

Solution:
#include <algorithm>
#include <cassert>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <ctime>
#include <deque>
#include <functional>
#include <iomanip>
#include <iostream>
#include <list>
#include <map>
#include <queue>
#include <random>
#include <set>
#include <sstream>
#include <string>
#include <utility>
#include <vector>

#define REP(i,s,n) for(int i=(int)(s);i<(int)(n);i++)

const int DEBUG = 0;


using namespace std;
typedef long long int ll;
typedef vector<int> VI;
typedef vector<ll> VL;
typedef pair<int, int> PI;
const ll mod = 1e9 + 7;

int n;

void ans_swap(VI &ans, int &pos, int j) {
  if (ans.size() > 100000) {
    assert (0);
  }
  while (pos != j + 1) {
    pos = (pos + 1) % n;
    ans.push_back(1);
  }
  ans.push_back(n - 1);
}

void shift(VI &ans, int &pos, int dest) {
  while (pos != dest) {
    pos = (pos + 1) % n;
    ans.push_back(1);
  }
}


int main(void) {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  VI p(n);
  REP(i, 0, n) {
    cin >> p[i];
  }
  VI cp(p);
  VI ans;
  int pos = 0;
  REP(i, 0, n - 1) {
    REP(j, 0, n - i - 1) {
      if (p[j] > p[j + 1]) {
	swap(p[j], p[j + 1]);
	ans_swap(ans, pos, j);
      }
    }
  }
  shift(ans, pos, 0);
  cout << ans.size() << "\n";
  REP(i, 0, ans.size()) {
    cout << ans[i] << "\n";
  }
  if (DEBUG) {
    REP(i, 0, ans.size()) {
      int k = ans[i];
      for(int i=k;i<n;i++)
	swap(cp[i],cp[i-k]);
    }
    REP(i, 0, n) {
      cerr << "result: ";
      for(auto cpp:cp)cerr << " " << cpp;
      cerr << endl;
    }
  }
}