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>
using namespace std;
#define int long long
#define vt vector
#define pb push_back
#define pob pop_back
#define vi vt<int>
#define vvi vt<vi>
#define vii vt<ii>
#define vvii vt<vii>
typedef long long ll;
#define For(i,a,b) for(int i = a; i < b; i++)
#define foR(i,a,b) for(int i = a; i <= b; i++)
#define roF(i,a,b) for(int i = a - 1; i >= b; i--)
#define Rof(i,a,b) for(int i = a; i >= b; i--)

int a[1000];
vi ans;
signed main (){
//	freopen("test.inp","r",stdin);
//	freopen(".out","w",stdout);
	int n;
	cin >> n;
	For(i,0,n){
		cin >> a[i];
	}
	Rof(i,n-1,0){
		int vt;
		For(j,0,n){
			if(a[j] == i){
				vt = j;
				break;
			}
		}
		if(vt == i){
			continue;
		}
		For(j,0,vt){
			ans.pb(1);
		}
		For(j,vt,i){
			ans.pb(1);
			ans.pb(n-1);		
		}
		For(j,i,n){
			ans.pb(1);
		}
		For(j,vt,i){
			swap(a[j],a[j+1]);
		}
	}
	cout << ans.size() << "\n";
	For(i,0,ans.size()){
		cout << ans[i] << "\n";
	}
}

