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 <algorithm>
#include <cassert>
#include <cstdio>
#include <vector>
#define repeat(i, n) for (int i = 0; (i) < int(n); ++(i))
#define repeat_from(i, m, n) for (int i = (m); (i) < int(n); ++(i))
#define whole(x) begin(x), end(x)
using namespace std;

int main() {
    // input
    int n; scanf("%d", &n);
    vector<int> p(n); repeat (i, n) scanf("%d", &p[i]);
    // solve
    vector<int> history;
    auto operate = [&](int k) {
        repeat_from (i, k, n) {
            swap(p[i], p[i - k]);
        }
        history.push_back(k);
    };
    while (p[0] != 0 or not is_sorted(whole(p))) {
        if (p[0] < p[n - 1] and p[0] != 0) {
            operate(n - 1);
        } else {
            operate(1);
        }
    }
    // output
    assert (is_sorted(whole(p)));
    assert (history.size() <= 100000);
    printf("%d\n", int(history.size()));
    for (int result_i : history) {
        printf("%d\n", result_i);
    }
    return 0;
}
