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 pb push_back
#define eb emplace_back
#define allv(V) ((V).begin()),((V).end())
#define sorv(V) sort(allv(V))
#define univ(V) (V).erase(unique(allv(V)),(V).end())
#define revv(V) reverse(allv(V))
#define sz(V) ((int)(V).size())
#define upmin(a,b) (a)=min((a),(b))
#define upmax(a,b) (a)=max((a),(b))
#define rb(x) ((x)&(-(x)))
#define INF (0x3f3f3f3f)
#define INFLL (0x3f3f3f3f3f3f3f3fll)
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;

const int MAXN = 205;

int A[MAXN];

vector<int> Ans;

int N;

int main() {
	ios::sync_with_stdio(false);

	cin >> N;
	for(int i = 1; i <= N; i++) cin >> A[i];
	
	for(int t = 0; t <= N; t++) {
		vector<int> V;
		for(int i = 2; i <= N; i++) {
			if(A[i-1] > A[i]) {
				swap(A[i-1], A[i]);
				V.eb(i-1);
			}
		}
		if(V.empty()) break;

		for(int t = 0; t < V[0]; t++) Ans.eb(1);
		Ans.eb(N-1);
		for(int i = 1; i < sz(V); i++) {
			for(int t = 0; t < V[i]-V[i-1]; t++) Ans.eb(1);
			Ans.eb(N-1);
		}
		for(int t = 0; t < N-V.back(); t++) Ans.eb(1);
	}

	printf("%d\n", sz(Ans));
	for(int v : Ans) printf("%d\n", v);
	return 0;
}