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 <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <deque>
#include <set>
#include <map>
#include <unordered_map>
#include <functional>
#include <cstring>
#include <cmath>
#include <ctime>
#include <cstdlib>

using namespace std;
typedef long long llong;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<llong, llong> pll;

int n;
int p[200];
int main() {
	ios_base::sync_with_stdio(false); cin.tie(0);
	cin >> n;
	for (int i = 0; i < n; ++i) {
        cin >> p[i];
	}
	vector<int> ans;
	for (int i = 1; i < n; ++i) {
        for (int j = 1; j < n; ++j) {
            ans.push_back(1);
            if (p[j - 1] > p[j]) {
                ans.push_back(n - 1);
                swap(p[j - 1], p[j]);
            }
        }
        ans.push_back(1);
	}
	printf("%d\n", (int)ans.size());
	for (int i : ans) {
        printf("%d\n", i);
	}
	return 0;
}
