Mike and strings 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, ..., s_n each consisting of lowercase English letters. In one move he can choose a string s_i, 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 s_i. 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

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： 你可以将任意字符串第一个字符移到字符串尾部，可以重复此过程，问最少多少步，使得所有字符串相等

#include <iostream>
#include <cstring>
#include <string>
using namespace std;

int get_step(string a, string b){
    int len = a.length();
    int ans = 0;
    for(int i = 0; i < len; i++){
        if(a == b)
            return ans;
        //string b(b, 1);
        string t = "";
        t += b[0];
        b.erase(0,1);
        b += t;
        //cout << i << " " << b << endl;
        ans++;
    }
    return -1;
}

int main(){
    int n;
    cin >> n;
    string str[55];
    for(int i = 1; i <= n; i++){
        cin >> str[i];
    }
    int num = 1234567;
    for(int i = 1; i <= n; i++){
        int cmp = num;
        num = 0;
        for(int j = 1; j <= n; j++){
            if(get_step(str[i], str[j]) == -1){
                cout << -1 << endl;
                return 0;
            }
            num += get_step(str[i], str[j]);
        }
        num = min(num,cmp);
    }
    cout << num << endl;
    return 0;
}