Codeforces 798B

B. Mike and strings

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Mike has n strings s1, s2, ..., sn each consisting of lowercase English letters. In one move he can choose a string si, erase the first character and append it to the end of the string. For example, if he has the string "coolmike", in one move he can transform it into the string "oolmikec".

Now Mike asks himself: what is minimal number of moves that he needs to do in order to make all the strings equal?

Input

The first line contains integer n (1 ≤ n ≤ 50) — the number of strings.

This is followed by n lines which contain a string each. The i-th line corresponding to string si. Lengths of strings are equal. Lengths of each string is positive and don't exceed 50.

Output

Print the minimal number of moves Mike needs in order to make all the strings equal or print  - 1 if there is no solution.

Examples

input

4
xzzwo
zwoxz
zzwox
xzzwo

output

5

input

2
molzv
lzvmo

output

2

input

3
kc
kc
kc

output

0

input

3
aa
aa
ab

output

-1
题目大意：略。
方法：把每一个字符串作为对象和其他字符串进行比较，记录操作数，求出最小操作数。
代码

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
using namespace std;
const int N=55;
char a[1][N];
void move1(int n)//这里不能写move,不然和库函数重名。c-free上编译结果正确，但cf上编译结果错误。 
{
    char s=a[0][0];
    for(int i=1;i<n;i++)
    {
        a[0][i-1]=a[0][i];
    }
    a[0][n-1]=s;
    a[0][n]='';
}
int main()
{
    char s[N][N];
    int n;
    scanf("%d",&n);
    getchar();
    for(int i=0;i<n;i++)
    {
        gets(s[i]);
    }
    int ans=0x3f3f3f3f;
    int flag=0;
    {
        for(int i=0;i<n;i++)
        {
            int cnt=0;
            for(int j=0;j<n;j++)
            {
                strcpy(a[0],s[j]);
                int k=0;
                while(strcmp(s[i],a[0])!=0) 
                {
                    move1(strlen(s[i]));
                    cnt++;
                    k++;
                    if(k>strlen(s[i]))//如果k大于串长度，说明两个串永远不可能相等，输出-1。 
                    {
                        i=n;
                        flag=1; 
                        break;    
                    }
                }
            }
            ans=min(ans,cnt);
        }
    } 
    if(flag)cout<<-1<<endl;
    else cout<<ans<<endl;
    return 0;
}