Problem: During the study of the Martians Petya clearly understood that the Martians are absolutely lazy. They like to sleep and don't like to wake up. 

Imagine a Martian who has exactly n eyes located in a row and numbered from the left to the right from 1 to n. When a Martian sleeps, he puts a patch on each eye (so that the Martian morning doesn't wake him up). The inner side of each patch has an uppercase Latin letter. So, when a Martian wakes up and opens all his eyes he sees a string s consisting of uppercase Latin letters. The string's length is n. 

"Ding dong!" — the alarm goes off. A Martian has already woken up but he hasn't opened any of his eyes. He feels that today is going to be a hard day, so he wants to open his eyes and see something good. The Martian considers only m Martian words beautiful. Besides, it is hard for him to open all eyes at once so early in the morning. So he opens two non-overlapping segments of consecutive eyes. More formally, the Martian chooses four numbers a, b, c, d, (1 ≤ a ≤ b < c ≤ d ≤ n) and opens all eyes with numbers i such that a ≤ i ≤ b or c ≤ i ≤ d. After the Martian opens the eyes he needs, he reads all the visible characters from the left to the right and thus, he sees some word.

Let's consider all different words the Martian can see in the morning. Your task is to find out how many beautiful words are among them.

Input

The first line contains a non-empty string s consisting of uppercase Latin letters. The strings' length is n (2 ≤ n ≤ 105). The second line contains an integer m (1 ≤ m ≤ 100) — the number of beautiful words. Next m lines contain the beautiful words pi, consisting of uppercase Latin letters. Their length is from 1 to 1000. All beautiful strings are pairwise different.

Output

Print the single integer — the number of different beautiful strings the Martian can see this morning.

Examples

Input

ABCBABA
2
BAAB
ABBA


Output

1

Note

Let's consider the sample test. There the Martian can get only the second beautiful string if he opens segments of eyes a = 1, b = 2 and c = 4, d = 5 or of he opens segments of eyes a = 1, b = 2 and c = 6, d = 7. 

Solution (in JAVA):

Solution:
import java.math.*;
import java.util.*;
import java.io.*;
public class Main {
    public static InputReader in = new InputReader(System.in);
    public static PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));

    public static void main(String[] args)
    {
        String s;
        s=in.next();
        int q;
        q=in.nextInt();
        SAM1 sam=new SAM1();
        sam.init(410000,26);
        for(int i=0;i<s.length();++i)
            sam.insert(s.charAt(i)-'A',i+1);
        int ans=0;
        boolean flag=false;
        sam.topo();
        for(int i=sam.cur-1;i>0;--i)
            sam.pool[i].pre.endpos=Math.max(sam.pool[i].endpos,sam.pool[i].pre.endpos);
        sam.root.firstpos=s.length()+1;
        sam.root.endpos=s.length()+1;
        while(q--!=0)
        {
            flag=false;
            String qs;
            qs=in.next();
            SAM_node p1,p2;
            if(qs.length()==1)
                continue;
            for(int i=0;i<qs.length();++i)
            {
                p1=sam.root;
                p2=sam.root;
                for(int j=0;j<=i;++j)
                {
                    if(p1.next[qs.charAt(j)-'A']!=null)
                        p1=p1.next[qs.charAt(j)-'A'];
                    else
                    {
                        p1=null;
                        break;
                    }
                }
                for(int j=i+1;j<qs.length();++j)
                {
                    if(p2.next[qs.charAt(j)-'A']!=null)
                        p2=p2.next[qs.charAt(j)-'A'];
                    else
                    {
                        p2=null;
                        break;
                    }
                }
                if(p1!=null && p2!=null)
                {
                    if(p2.endpos-qs.length()+i>=p1.firstpos)
                        flag=true;
                }
            }
            if(flag)
            {
                ans++;
            }
        }
        out.println(ans);
        out.flush();
        out.close();
    }
}

class SAM_node
{
    SAM_node pre,next[];
    int step,cnt,firstpos,ans,endpos;
    SAM_node(int sigma)
    {
        endpos=0;
        next=new SAM_node[sigma];
        step=0;
        cnt=0;
        pre=null;
        firstpos=0;
        ans=0;
    }
}
class SAM1
{
    SAM_node SAM_pool[],root,last;
    int d[];
    SAM_node pool[];
    int cur;
    int sigma;
    void topo() {
        // 求出parent树的拓扑序
        int cnt = cur;
        int maxVal = 0;
        Arrays.fill(d, 0);
        for (int i = 1; i < cnt; i++) {
            maxVal = Math.max(maxVal, SAM_pool[i].step);
            d[SAM_pool[i].step]++;
        }
        for (int i = 1; i <= maxVal; i++)
            d[i] += d[i - 1];
        for (int i = 1; i < cnt; i++)
            pool[d[SAM_pool[i].step]--] = SAM_pool[i];
        pool[0] = root;
    }
    void init(int a,int b)
    {
        d=new int[a];
        pool=new SAM_node[a];
        SAM_pool=new SAM_node[a];
        SAM_pool[0]=new SAM_node(b);
        sigma=b;
        root=last=SAM_pool[0];
        cur=1;
    }
    void insert(int w,int L)
    {
        SAM_node p=last;
        SAM_pool[cur]=new SAM_node(sigma);
        SAM_node np=SAM_pool[cur];
        last=np;
        cur++;
        np.step=p.step+1;
        np.firstpos = np.step; //确定原串初始位置
        while (p!=null && p.next[w]==null)
        {
            p.next[w]=np;
            p = p.pre;
        }
        if(p==null)
        {
            np.pre=root;
        }
        else
        {
            SAM_node q=p.next[w];
            if(p.step+1==q.step)
                np.pre=q;
            else {
                SAM_node nq = SAM_pool[cur++] = new SAM_node(sigma);
                nq.next = Arrays.copyOf(q.next, sigma);
                nq.firstpos = q.firstpos;
                nq.step = p.step + 1;
                nq.pre = q.pre;
                q.pre = nq;
                np.pre = nq;
                while (p != null && p.next[w]==(q)) {
                    p.next[w] = nq;
                    p = p.pre;
                }
            }
        }
        np.endpos=L;
    }
}
class InputReader{
    private final static int BUF_SZ = 65536;
    BufferedReader in;
    StringTokenizer tokenizer;
    public InputReader(InputStream in) {
        super();
        this.in = new BufferedReader(new InputStreamReader(in),BUF_SZ);
        tokenizer = new StringTokenizer("");
    }
    public String next() {
        while (!tokenizer.hasMoreTokens()) {
            try {
                tokenizer = new StringTokenizer(in.readLine());
            } catch (IOException e) {
                throw new RuntimeException(e);
            }
        }
        return tokenizer.nextToken();
    }
    public int nextInt() {
        return Integer.parseInt(next());
    }
    public long nextLong()
    {
        return Long.parseLong(next());
    }
}