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;
typedef long long ll;

#define EPS (1e-7)
#define INF (1e9)
#define PI (acos(-1))
//const ll mod = 1000000007;
int N;
int a[205];

bool is_sorted() {
    for(int i = 1; i < N; i++) {
        if(a[i - 1] >= a[i]) return false;
    }
    return true;
}
vector<int> ans;

int main() {
    //cout.precision(10);
    cin >> N;
    for(int i = 0; i < N; i++) {
        cin >> a[i];
    }
    while(!is_sorted()) {
        for(int i = 0; i < N; i++) {
            //cout << a[i] << " ";
        }
        //cout << endl;
        //cout << 1 << endl;
        ans.push_back(1);
        int tmp = a[0];
        for(int i = 0; i < N - 1; i++) {
            a[i] = a[i + 1];
        }
        a[N - 1] = tmp;
        if(is_sorted()) break;
        if(a[0] < a[N - 1]) {
            if(a[0] == 0 && a[N - 1] == (N - 1)) continue;
            //cout << N - 1 << endl;
            ans.push_back(N - 1);
            swap(a[0], a[N - 1]);
        }
    }
    cout << ans.size() << endl;
    for(int i = 0; i < ans.size(); i++) {
        cout << ans[i] << endl;
    }
    return 0;
}