Problem:
Score : 900 points

Problem Statement

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).

Partial Scores

In the test set worth 300 points, N \leq 7.

In the test set worth another 400 points, N \leq 30.

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.

Sample Input 1

5
4 2 0 1 3

Sample Output 1

4
2
3
1
2

Each operation changes p as shown below:

Sample Input 2

9
1 0 4 3 5 6 2 8 7

Sample Output 2

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

Solution in C++14 (GCC 5.4.1) (judge verdict: Runtime Error)

#define _CRT_SECURE_NO_WARNINGS
#define _SCL_SECURE_NO_WARNINGS
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <string>
#include <vector>
#include <list>
#include <utility>
#include <algorithm>
#include <functional>
#include <cmath>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <iomanip>
#include <sstream>
#include <bitset>
#include <limits>
#include <numeric>
#include <valarray>

using namespace std;
typedef unsigned int uint;
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<LL, LL> PP;
#define REP(i,a,n) for(LL i = (a); i < (LL)(n); ++i)
#define REM(i,a,n) for(LL i = ((n) - 1); i >= (a); --i)
#define FLOAT setprecision(16)
#define SPEEDUP {cin.tie(0); ios::sync_with_stdio(false);}
const int INF = 0x3FFFFFFF;
const LL INFLL = 0x3FFFFFFF3FFFFFFF;
const double INFD = 1.0e+308;
const string INFSTR = "\x7f";
const double EPS = 1.0e-9;

void YesNo(bool b) { cout << (b ? "Yes" : "No") << endl; }
void YESNO(bool b) { cout << (b ? "YES" : "NO") << endl; }
template <class T, class U>
istream& operator>>(istream& ist, pair<T, U>& right) { return ist >> right.first >> right.second; }
template <class T, class U>
ostream& operator<<(ostream& ost, const pair<T, U>& right) { return ost << right.first << ' ' << right.second; }
template <class T, class TCompatible, size_t N>
void Fill(T(&dest)[N], const TCompatible& val) { fill(dest, dest + N, val); }
template <class T, class TCompatible, size_t M, size_t N>
void Fill(T(&dest)[M][N], const TCompatible& val) { for (int i = 0; i < M; ++i) Fill(dest[i], val); }
template<class T>
T Compare(T left, T right) { return left > right ? 1 : (left < right ? -1 : 0); }
istream& Ignore(istream& ist) { string s; ist >> s; return ist; }
bool Inside(int i, int j, int h, int w) { return i >= 0 && i < h && j >= 0 && j < w; }
// all_of
// partial_sum, adjacent_difference

#ifdef ONLY_MY_ENVIR
#include "IntMod.h"
#include "Union_Find.h"
#include "Graph.h"
#include "Range.h"
#include "Global.h"
#include "Flow_Solver.h"
#include "Tree.h"
#include "Suffix_Array.h"
#include "Geometry.h"
#include "Matrix.h"
#endif

#ifdef __GNUC__
typedef __int128 LLL;
istream& operator>> (istream& ist, __int128& val) { LL tmp;  ist >> tmp; val = tmp; return ist; }
ostream& operator<< (ostream& ost, __int128 val) { LL tmp = val; ost << tmp; return ost; }
#endif

#if 1234567891
#include <array>
#include <random>
#include <unordered_set>
#include <unordered_map>
template<typename T>
using PriorityQ = priority_queue<T, vector<T>, greater<T> >;	// コスト小を優先
#endif

int N;
vector<int> ans;
int main() {
	cin >> N;
	deque<int> D, R(N);
	iota(R.begin(), R.end(), 0);
	REP(i, 0, N) {
		int a;
		cin >> a;
		D.push_back(a);
	}

	if (N % 2 == 0) throw 1;
	while (D != R) {
		int a = D.front(); D.pop_front();
		int b = D.front(); D.pop_front();
		if (a < b) {
			D.push_back(a);
			D.push_back(b);
			ans.push_back(1);
		} else {
			D.push_back(b);
			D.push_back(a);
			ans.push_back(2);
		}
	}
	cout << ans.size() << endl;
	for (int i : ans) {
		cout << i << endl;
	}

	return 0;
}