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 VARNAME(x) #x
#define show(x) cerr << #x << " = " << x << endl

using namespace std;
using ll = long long;

template <typename T>
ostream& operator<<(ostream& os, const vector<T>& v)
{
    os << "sz:" << v.size() << "\n[";
    for (const auto& p : v) {
        os << p << ",";
    }
    os << "]\n";
    return os;
}

template <typename S, typename T>
ostream& operator<<(ostream& os, const pair<S, T>& p)
{
    os << "(" << p.first << "," << p.second
       << ")";
    return os;
}


constexpr ll MOD = (ll)1e9 + 7LL;

template <typename T>
constexpr T INF = numeric_limits<T>::max() / 10;

int N;
void query(vector<int>& p, const int k, vector<int>& op)
{
    op.push_back(k);
    for (int i = k; i < N; i++) {
        swap(p[i], p[i - k]);
    }
}

int main()
{
    cin.tie(0);
    ios::sync_with_stdio(false);

    cin >> N;
    vector<int> p(N);
    for (int i = 0; i < N; i++) {
        cin >> p[i];
    }

    vector<int> op;
    for (int i = 0; i < 2 * N * N; i++) {
        if (p[0] > 0 and p[N - 1] > 0 and p[0] < p[N - 1]) {
            query(p, N - 1, op);
        }
        query(p, 1, op);
    }
    while (p[0] > 0) {
        query(p, 1, op);
    }

    cout << op.size() << endl;
    for (const int o : op) {
        cout << o << endl;
    }
    show(p);

    return 0;
}
