B. Obtaining the String

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given two strings s

and t. Both strings have length n and consist of lowercase Latin letters. The characters in the strings are numbered from 1 to n

.

You can successively perform the following move any number of times (possibly, zero):

• swap any two adjacent (neighboring) characters of s

(i.e. for any i={1,2,…,n−1} you can swap si and si+1)

• .

You can't apply a move to the string t

. The moves are applied to the string s

one after another.

Your task is to obtain the string t

from the string s. Find any way to do it with at most 104

such moves.

You do not have to minimize the number of moves, just find any sequence of moves of length 104

or less to transform s into t

.

Input

The first line of the input contains one integer n

(1≤n≤50) — the length of strings s and t

.

The second line of the input contains the string s

consisting of n

lowercase Latin letters.

The third line of the input contains the string t

consisting of n

lowercase Latin letters.

Output

If it is impossible to obtain the string t

using moves, print "-1".

Otherwise in the first line print one integer k

— the number of moves to transform s to t. Note that k must be an integer number between 0 and 104

inclusive.

In the second line print k

integers cj (1≤cj<n), where cj means that on the j-th move you swap characters scj and scj+1

.

If you do not need to apply any moves, print a single integer 0

in the first line and either leave the second line empty or do not print it at all.

Examples

Input

Copy

6
abcdef
abdfec

Output

Copy

4
3 5 4 5 

Input

Copy

4
abcd
accd

Output

Copy

-1

Note

In the first example the string s

changes as follows: "abcdef" → "abdcef" → "abdcfe" → "abdfce" →

"abdfec".

In the second example there is no way to transform the string s

into the string t through any allowed moves.

題意：

給你兩個字串a， b和它們的長度， 每次只能移動前後兩個字母， 求字串a最少需要移動多少次才能變成字串b。
當時做這個題的時候可把我嚇壞了。然後我就把這題跳過去了。 補題的時候看了個學長的程式碼發現如此簡單， 果然還是自己太笨了。

（ 1）先比較兩個字串的大小， 如果不相等的話再怎麼移也變不成b。

（ 2）再接下來就是模擬過程了， 從頭挨個字串比較， 直到不相等的時候， 然後找從a中找當前與b[i]相等的值， 然後將它移動到a[i]這個位置， 記錄下步數。 然後以此類推。

程式碼如下：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <string>
#include <queue>
//1584
using namespace std;
char a[50],b[50];
char aa[50],bb[50];
int main()
{
    int n;
    scanf("%d",&n);
    scanf("%s\n%s",a,b);
    strcpy(aa,a);
    strcpy(bb,b);
    sort (aa,aa+n);
    sort (bb,bb+n);
    queue<int>q;
    if(strcmp(aa,bb)!=0)
        printf("-1\n");
    else
    {
        for (int i=0;i<n;i++)
        {
            if(a[i]==b[i])
               continue;
                for (int j=i;j<n;j++)
                {
                    if(b[i]==a[j])
                    {
                        for (int k=j;k>i;k--)
                            {
                                  q.push(k);
                                  swap(a[k],a[k-1]);
                            }break;
                    }

                }
        }
        printf("%d\n",q.size());
        while (!q.empty())
        {
            printf("%d ",q.front());
            q.pop();
        }
        printf("\n");
    }

    return 0;
}