【CodeFores 798B】 Mike and strings（模擬+string）



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".

題目大意：給n個字符串，將第i個字符串移動xi位，使得最終所有字符串相等，求最小的移動步數和，若無法相等輸出-1

思路：n最大50個，枚舉每一個字符串作爲基礎串，求得每種情況的移動步數和，取最小值。string find大發好，比賽時忘了用這個，結果沒懟出來。果然還是不太熟悉

#include <bits/stdc++.h>
#define INF 0x3f3f3f3f
using namespace std;

int main()
{
    int n,ans;
    string s,t,p;
    while(~scanf("%d",&n)){
        vector<string>v;
        ans=INF;
        for (int i=0; i<n; i++){
            cin>>s;
            v.push_back(s);
        }
        for (int i=0; i<n; i++){
            int sum=0;
            t=v[i];
            for (int j=0; j<n; j++){
                if (i!=j){
                    p=v[j];
                    p+=p;
                    int k=p.find(t);   //返回下標
                    if (k==-1){
                        sum=-1;
                        break;
                    }
                    sum+=k;
                }
            }
            if (sum != -1) ans=min(ans,sum);
        }
        if (ans==INF) printf("-1\n");
        else printf("%d\n",ans);
    }
    return 0;
}