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.io.*;
import java.util.*;
import static java.lang.Math.*;
import static java.util.Arrays.*;
import static java.util.Collections.*;
public class Main{
    public static void main(String[]args){
        InputStream in = System.in;
        PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));
        try{
            //          in = new BufferedInputStream(new FileInputStream(new File("input.txt")));
            //          out = new PrintWriter(new BufferedOutputStream(new FileOutputStream("output.txt")));
        }catch (Exception e) {
        }
        new Main().run(in,out);
        out.flush();
    }
    private void debug(Object...os){
        System.err.println(deepToString(os));
    }
    
    private void run(InputStream in,PrintWriter out){
        String s=next(in);
        String rs=new StringBuilder(s).reverse().toString();
        int m=nextInt(in);
        int res=0;
        for (int i = 0; i < m; i++) {
            String t=next(in);
            if(t.length()<=1){
                continue;
            }
            int[] next=generateFailureTable(t);
            String r=new StringBuilder(t).reverse().toString();
            int[] rnext=generateFailureTable(r);
            int[] minIndex=new int[t.length()+1];
            fill(minIndex,-1);
            int p=0;
            for (int j = 0; j < s.length() && p<t.length(); j++) {
                while(p>=0 && s.charAt(j)!=t.charAt(p))p=next[p];
                p++;
                if(minIndex[p]==-1)minIndex[p]=j+1;
            }
            int[] rminIndex=new int[r.length()+1];
            fill(rminIndex,-1);
            p=0;
            for (int j = 0; j < rs.length() && p<r.length(); j++) {
                while(p>=0 && rs.charAt(j)!=r.charAt(p))p=rnext[p];
                p++;
                if(rminIndex[p]==-1)rminIndex[p]=j+1;
            }
            if(minIndex[t.length()]>=0){
                res++;
            }else{
                for (int j = 0; j < minIndex.length; j++) {
                    int k=t.length()-j;
                    if(minIndex[j]>=0 && rminIndex[k]>=0 && (minIndex[j]+rminIndex[k])<=s.length()){
                        res++;break;
                    }
                }
            }
        }
        out.println(res);
    }
    
    /**
     * pattern = sに対する,KMP法で用いられる,failureTable を作る.
     * table[i] は, s[i] と,比較文字列 t[j] が異なった場合に,
     * 次は, s[table[i]] と, t[j] を比較すべきことを表す.
     * 
     * table[0] = -1 であり,この場合は, j++ する.
     * @param pattern
     * @return
     */
    static int[] generateFailureTable(String pattern){
        int[]next=new int[pattern.length()];
        next[0] = -1;
        for (int i = 1, j = -1; i < pattern.length(); i++) {
            while(j>=0 && pattern.charAt(i-1) != pattern.charAt(j)){
                j = next[j];
            }
            next[i]=++j;
        }
        return next;
    }
    private String nextLine(InputStream in) {
        try {
            StringBuilder res = new StringBuilder("");
            int c = in.read();
            if(c==-1)return null;
            while (c=='\n' || c=='\r'){
                c = in.read();
                if(c==-1)return null;
            }
            do {
                res.append((char) c);
                c=in.read();
            } while (c!='\n' && c!='\r');
            return res.toString();
        } catch (Exception e) {
            return null;
        }
    }
    
    private String next(InputStream in) {
        try {
            StringBuilder res = new StringBuilder("");
            int c = in.read();
            if(c==-1)return null;
            while (Character.isWhitespace(c)){
                c = in.read();
                if(c==-1)return null;
            }
            do {
                res.append((char) c);
            } while (!Character.isWhitespace(c = in.read()));
            return res.toString();
        } catch (Exception e) {
            return null;
        }
    }
    
    private int nextInt(InputStream in){
        try{
            int c=in.read();
            if(c==-1) return c;
            while(c!='-'&&(c<'0'||'9'<c)){
                c=in.read();
                if(c==-1) return c;
            }
            if(c=='-') return -nextInt(in);
            int res=0;
            do{
                res*=10;
                res+=c-'0';
                c=in.read();
            }while('0'<=c&&c<='9');
            return res;
        }catch(Exception e){
            return -1;
        }
    }
    
    private long nextLong(InputStream in){
        try{
            int c=in.read();
            if(c==-1) return c;
            while(c!='-'&&(c<'0'||'9'<c)){
                c=in.read();
                if(c==-1) return c;
            }
            if(c=='-') return -nextLong(in);
            long res=0;
            do{
                res*=10;
                res+=c-'0';
                c=in.read();
            }while('0'<=c&&c<='9');
            return res;
        }catch(Exception e){
            return -1;
        }
    }
}