B. Obtaining the String

本文介绍了一个算法,用于解决通过交换相邻字符的方式将一个字符串转换为另一个字符串的问题。输入包含两个长度相同的字符串,目标是在限定步数内完成转换。文章提供了一种实现思路及代码示例,演示如何有效进行字符串匹配与转换。

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http://codeforces.com/contest/1015/problem/B

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.

 

 

 


#include<bits/stdc++.h>
using namespace std;
int main()
{
    int n;
    string a,b;
    int x[10009];
    while(scanf("%d",&n)!=EOF)
    {
        memset(x,0,sizeof(x));
        int t=0,T=1;
        cin>>a;
        cin>>b;
        char a1[55],b1[55];
        for(int i=0;i<n;i++)a1[i]=a[i],b1[i]=b[i];
        sort(a1,a1+n);
        sort(b1,b1+n);
        for(int i=0;i<n;i++)
        {
            if(a1[i]!=b1[i])
            {
                T=0;break;
            }
        }
        if(T==0)
            cout<<"-1"<<endl;
        else
        {
            for(int i=0;i<n;i++)
            {
                if(a[i]!=b[i])
                {
                    int j=i;
                    for( ;j<n;j++)
                    {
                        if(a[j]==b[i])
                            break;
                    }
                    for(int k=j;k>i;k--)
                        {
                            char g;
                            g=a[k];
                            a[k]=a[k-1];
                            a[k-1]=g;
                            x[++t]=k;
                            }
                }
            }
            cout<<t<<endl;
            for(int j=1;j<=t;j++)
                cout<<x[j]<<" ";
            cout<<endl;
        }
    }
    return 0;
}

 

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