Codeforces Round #394 (Div. 2) C. Dasha and Password（简单DP）

C. Dasha and Password

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

After overcoming the stairs Dasha came to classes. She needed to write a password to begin her classes. The password is a string of length n which satisfies the following requirements:

• There is at least one digit in the string,
• There is at least one lowercase (small) letter of the Latin alphabet in the string,
• There is at least one of three listed symbols in the string: '#', '*', '&'.

Considering that these are programming classes it is not easy to write the password.

For each character of the password we have a fixed string of length m, on each of these n strings there is a pointer on some character. The i-th character displayed on the screen is the pointed character in the i-th string. Initially, all pointers are on characters with indexes 1 in the corresponding strings (all positions are numbered starting from one).

During one operation Dasha can move a pointer in one string one character to the left or to the right. Strings are cyclic, it means that when we move the pointer which is on the character with index 1 to the left, it moves to the character with the index m, and when we move it to the right from the position m it moves to the position 1.

You need to determine the minimum number of operations necessary to make the string displayed on the screen a valid password.

Input

The first line contains two integers n, m (3 ≤ n ≤ 50, 1 ≤ m ≤ 50) — the length of the password and the length of strings which are assigned to password symbols.

Each of the next n lines contains the string which is assigned to the i-th symbol of the password string. Its length is m, it consists of digits, lowercase English letters, and characters '#', '*' or '&'.

You have such input data that you can always get a valid password.

Output

Print one integer — the minimum number of operations which is necessary to make the string, which is displayed on the screen, a valid password.

Examples

Input

3 4
1**2
a3*0
c4**

Output

1

Input

5 5
#*&#*
*a1c&
&q2w*
#a3c#
*&#*&

Output

3

Note

In the first test it is necessary to move the pointer of the third string to one left to get the optimal answer.

【分析】题意很简单，就是定义一个合格的密码，必须包括数字（0~9），字母，符号这三样，然后给你一个二维字符串，在每一行选一个字符，使得每一行选的字符组合起来的字符串为合格的密码。一开始每一行的指针都在第一列，然后每次只能向左右移动一步，若在最左边，向左移可到达最右边，在最右边向右移则到达最左边。问最小步骤。简单DP一下就行了。

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <string>
#include <stack>
#include <queue>
#include <vector>
#define inf 0x3f3f3f3f
#define met(a,b) memset(a,b,sizeof a)
#define pb push_back
#define inf 0x3f3f3f3f
using namespace std;
typedef long long ll;
const int N = 35;
const int M = 1e5+10;
int n,m,ans=12345678;
char s[105][105];
int dist[105][5];
int main()
{
    int i,j,k;
    scanf ("%d%d",&n,&m);
    for(int i=0;i<105;i++)for(int j=0;j<5;j++)dist[i][j]=12345678;
    for (i=1;i<=n;i++)
        for (j=1;j<=m;j++)
            scanf (" %c",&s[i][j]);
    for (i=1;i<=n;i++)
    {    for (j=1;j<=m;j++)
            if (s[i][j]>='0'&&s[i][j]<='9') dist[i][1]=min(dist[i][1],min(j-1,m-j+1));
            else if (s[i][j]>='a'&&s[i][j]<='z') dist[i][2]=min(dist[i][2],min(j-1,m-j+1));
            else dist[i][3]=min(dist[i][3],min(j-1,m-j+1));
    }
    for (i=1;i<=n;i++)
        for (j=1;j<=n;j++) if (j!=i)
            for (k=1;k<=n;k++) if (k!=i&&k!=j)
                ans=min(ans,dist[i][1]+dist[j][2]+dist[k][3]);
    cout<<ans<<endl;
    return 0;
}