CF798B：Mike and strings（Hash）

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

4xzzwozwoxzzzwoxxzzwo

output

5

input

2molzvlzvmo

output

2

input

3kckckc

output

0

input

3aaaaab

output

-1

Note

In the first sample testcase the optimal scenario is to perform operations in such a way as to transform all strings into "zwoxz".

题意：给N个字符串，每次可以选择一个串循环左移一位，问最少操作几次使得所有串相等。

思路：范围不大，每个串每个位置为起点记录哈希值及它移到开头的次数。

# include <bits/stdc++.h>using namespace std;typedef unsigned long long ull;map<ull, int>m, o_o;int main(){    int n;    char s[53];    scanf("%d",&n);    for(int _=0; _<n; ++_)    {        scanf("%s",s);        int len = strlen(s);        o_o.clear();        for(int i=0; i<len; ++i)        {            ull h = s[i];            for(int j=(i+1)%len;j!=i; j=(j+1)%len)                h = h*3+s[j];            if(_ && !m.count(h)) return 0*puts("-1");            if(!o_o.count(h)) m[h] += i;            ++o_o[h];        }    }    int ans = 0x3f3f3f3f;    for(auto it = m.begin(); it!=m.end(); ++it)        ans = min(ans, it->second);    printf("%d\n",ans);    return 0;}