ZCMU2204 Keyboard(vector应用)

Description

 Vasya learns to type. He has an unusual keyboard at his disposal: it is rectangular and it has n rows of keys containing m keys in each row. Besides, the keys are of two types. Some of the keys have lowercase Latin letters on them and some of the keys work like the "Shift" key on standard keyboards, that is, they make lowercase letters uppercase.

 Vasya can press one or two keys with one hand. However, he can only press two keys if the Euclidean distance between the centers of the keys does not exceed x. The keys are considered as squares with a side equal to 1. There are no empty spaces between neighbouring keys.

 Vasya is a very lazy boy, that's why he tries to type with one hand as he eats chips with his other one. However, it is possible that some symbol can't be typed with one hand only, because the distance between it and the closest "Shift" key is strictly larger than x. In this case he will have to use his other hand. Having typed the symbol, Vasya returns other hand back to the chips.

 You are given Vasya's keyboard and the text. Count the minimum number of times Vasya will have to use the other hand.

Input

 The first line contains three integers n, m, x (1≤n,m≤30,1≤x≤50).

 Next n lines contain descriptions of all the keyboard keys. Each line contains the descriptions of exactly m keys, without spaces. The letter keys are marked with the corresponding lowercase letters. The "Shift" keys are marked with the "S" symbol.

 Then follow the length of the text q (1≤q≤5·105). The last line contains the text T, which consists of q symbols, which are uppercase and lowercase Latin letters.

Output

If Vasya can type the text, then print the minimum number of times he will have to use his other hand. Otherwise, print "-1" (without the quotes).

Examples

Input

2 2 1
ab
cd
1
A

Output

-1

Input

2 2 1
ab
cd
1
e

Output

-1

Input

2 2 1
ab
cS
5
abcBA

Output

1

Input

3 9 4
qwertyuio
asdfghjkl
SzxcvbnmS
35
TheQuIcKbRoWnFOXjummsovertHeLazYDOG

Output

2

Note

In the first sample the symbol "A" is impossible to print as there's no "Shift" key on the keyboard.

In the second sample the symbol "e" is impossible to print as there's no such key on the keyboard.

In the fourth sample the symbols "T", "G" are impossible to print with one hand. The other letters that are on the keyboard can be printed. Those symbols come up in the text twice, thus, the answer is 2.

思路

无法敲出所给文本有两种情况:

1.文本中存在键盘中不存在的字符

2.文本中有大写字符但键盘不存在shift键

至于如何得出用双手操作的次数我们可以先进行一些处理: 记录键盘中字符的位置(每种字符的个数不一定只有一个)

之后对文本进行遍历，一 一计算其对应的小写字符到'S'的距离是否小于x

代码

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<iostream>
#include<vector>
using namespace std;
#define MAX 500005
int le[26],flag;
struct node
{
    int x;
    int y;
};
vector<node> e[26];
vector<node> s;
void init()
{
    for(int i=0;i<26;i++)
        e[i].clear();
    s.clear();
}
int check(char *str,int l)
{
    for(int i=0; i<l; i++)
    {
        if(str[i]>='A'&&str[i]<='Z')
        {
            char temp=str[i]-'A';
            if(!flag||!le[temp])
                return 0;
        }
        if(str[i]>='a'&&str[i]<='z')
        {
            if(!le[str[i]-'a'])
                return 0;
        }
    }
    return 1;
}
int solve(char *str,int l,double step)
{
    int time=0;
    for(int i=0; i<l; i++)
    {
        if(str[i]>='A'&&str[i]<='Z')
        {
            char temp=str[i]-'A';
            int mark=0;
            for(int j=0; j<e[temp].size(); j++)
            {
                if(mark==1) break;
                for(int k=0; k<s.size(); k++)
                {
                    double dis=sqrt(1.0*(e[temp][j].x-s[k].x)*(e[temp][j].x-s[k].x)+1.0*(e[temp][j].y-s[k].y)*(e[temp][j].y-s[k].y));
                    if(dis<=step)
                    {
                        mark=1;
                        break;
                    }
                }
            }
            if(!mark) time+=1;
        }
    }
    return time;
}
int main()
{
    int n,m;
    double x;
    while(~scanf("%d %d %lf",&n,&m,&x))
    {
        char grid[50][50],str[MAX];
        memset(le,0,sizeof(le));
        init();
        for(int i=0; i<n; i++)
        {
            scanf("%s",grid[i]);
        }
        flag=0;
        for(int i=0; i<n; i++)
        {
            for(int j=0; j<m; j++)
            {
                if(grid[i][j]>='a'&&grid[i][j]<='z')
                {
                    le[grid[i][j]-'a']=1;
                    node temp;
                    temp.x=i; temp.y=j;
                    e[grid[i][j]-'a'].push_back(temp);
                }
                else if(grid[i][j]=='S')
                {
                    node temp;
                    temp.x=i; temp.y=j;
                    s.push_back(temp);
                    flag=1;
                }
            }
        }
        int l,mark=0;
        scanf("%d %s",&l,str);
        if(!check(str,l))
            printf("-1\n");
        else
            printf("%d\n",solve(str,l,x));
    }
    return 0;
}