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
#define  _CRT_SECURE_NO_WARNINGS
#include <stdio.h>
#include <algorithm>
#include <utility>
#include <functional>
#include <cstring>
#include <queue>
#include <stack>
#include <math.h>
#include <iterator>
#include <vector>
#include <string>
#include <set>
#include <math.h>
#include <iostream>
#include <random>
#include<map>
#include <iomanip>
#include <time.h>
#include <stdlib.h>
#include <list>
#include <typeinfo>
#include <list>
#include <set>
#include <cassert>
#include<fstream>
#include <unordered_map>
#include <cstdlib>
#include <complex>
#include <cctype>
#include <bitset>
using namespace std;
typedef string::const_iterator State;
#define Ma_PI 3.141592653589793
#define eps 1e-5
#define LONG_INF 1e18
#define GOLD 1.61803398874989484820458
#define MAX_MOD 1000000007LL
#define GYAKU 500000004LL
#define MOD 998244353LL
#define seg_size 262144*4
#define REP(a,b) for(long long a = 0;a < b;++a)
int gogo[100000];
int main(){
	int n;
	cin >> n;
	REP(i, n) {
		cin >> gogo[i];
	}
	vector<int> answer;
	for (int i = 0;; ++i) {
		long long geko = (i + n - 1) % n;
		if (geko != n - 1) {
			if (gogo[geko] > gogo[i%n]) {
				swap(gogo[i%n], gogo[geko]);
				answer.push_back(n - 1);
			}
		}
		int ok = 1;
		for (int q = 0; q < n; ++q) {
			if (gogo[q] != q) {
				ok = 0;
				break;
			}
		}
		if (ok == 1&&i % n == 0) {
			cout << answer.size() << endl;
			REP(q, answer.size()) {
				cout << answer[q] << endl;
			}
			return 0;
		}
		answer.push_back(1);
	}
}