B. Mike and Strings

B. Mike and strings time limit per test 2 seconds memory limit per test megabytes input standard input output standard Output

Mike has n strings s1, s2, ..., sn each consisting of lowercase 中文版 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 had the string "Coolmike", in one move he can transform it into the string "Oolmikec".

Now Mike asks Himself:what are minimal number of moves that he needs to does 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 are 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

Note

In the first sample TestCase the optimal scenario are to perform operations in such a-to-transform all-strings into "Zwoxz".

At first, did not consider the situation of the circulation section, and then the direct violence on the line Qaq,:-(is a wave off points, fast on 1200 qaq.

#include <bits/stdc++.h>

using namespace std;
const int maxn=3e5+7;
const int INF=1E9;
Char s[60][60];
Char s1[60];
int l;
int n;
int num[60];
int check (char *s)
{for
    (int i=0; i<l; ++i)
    {
        int J;
        for (j=0; j<l; ++j)
        {
            if (s1[j]!=s[(j+i)%l]) break;
        }
        if (j>=l) return i;
    }
    return-1;
}
int main ()
{
    scanf ("%d", &n);
    for (int i=0; i<n; ++i) scanf ("%s", S[i]);
    L=strlen (S[0]);
    int ans=inf;
    for (int i=0; i<l; ++i)
    {
        int sum=0;
        for (int j=0; j<l; ++j) s1[j]=s[0][(j+i)%l];
        for (int j=0; j<n; ++j)
        {
            int t=check (s[j]);
            if (t==-1)
            {
                puts ("-1");
                return 0;
            }
            sum+=t;
        }
        Ans=min (ans,sum);
    }
    printf ("%d\n", ans);
    return 0;
}