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
#include <bits/stdc++.h>
#define xx first
#define yy second
#define mp make_pair
#define pb push_back
#define mset(x, y) memset(x, y, sizeof x)
#define mcpy(x, y) memcpy(x, y, sizeof x)
using namespace std;

typedef long long LL;
typedef pair <int, int> pii;

inline int Read()
{
	int x = 0, f = 1, c = getchar();
	for (; !isdigit(c); c = getchar())
		if (c == '-')
			f = -1;
	for (;  isdigit(c); c = getchar())
		x = x * 10 + c - '0';
	return x * f;
}

const int MAXN = 205;

vector <int> ans;
int n, a[MAXN];

inline void Solve(int x)
{
	ans.pb(x);
	for (int i = x; i < n; i ++)
		swap(a[i], a[i - x]);
}

int main()
{
#ifdef wxh010910
	freopen("data.in", "r", stdin);
#endif
	n = Read();
	for (int i = 0; i < n; i ++)
		a[i] = Read();
	while (!is_sorted(a, a + n))
		if (a[0] && a[0] < a[n - 1])
			Solve(n - 1);
		else
			Solve(1);
	printf("%d\n", ans.size());
	for (int i = 0; i < ans.size(); i ++)
		printf("%d\n", ans[i]);
	return 0;
}
