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.ByteArrayInputStream;
import java.io.File;
import java.io.FileInputStream;
import java.io.FileNotFoundException;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.nio.charset.StandardCharsets;
import java.util.InputMismatchException;

public class E149 {
/*
ABCBABA
2
BAAB
ABBA

BALBABATLBAT
3
BALBAT
ALBALBA
LBLAB

 */
	static class Solver {
		
		int N, K, eyelid[];
		int space[] = new int[110_000], spaceRev[] = new int[110_000];
		int pref[] = new int[110_000], prefRev[] = new int[110_000];
		
		void solve(FastScanner s, PrintWriter out) {
			N = (eyelid = s.next().chars().toArray()).length;
			K = s.nextInt();
			
			int tot = 0, word[], M;
			www: while(K --> 0) {
				M = (word = s.next().chars().toArray()).length;
				if(M == 1) continue;
				// space is word # eyelid
				// spRev is drow # dileye
				for(int i = 0; i < M; i++)
					space[i] = spaceRev[M - 1 - i] = word[i];
				space[M] = spaceRev[M] = -1;
				for(int i = 0; i < N; i++)
					space[M + 1 + i] = spaceRev[M + N - i] = eyelid[i];
				kmp(space, pref, M + N + 1);
				kmp(spaceRev, prefRev, M + N + 1);
				int mx = 0;
				for(int i = 0; i < N; i++) {
					mx = max(mx, prefRev[M + 1 + i]);
					prefRev[M + 1 + i] = mx;
				}
				for(int i = 0; i < N; i++) {
					int cur = pref[M + 1 + i];
					int ofc = prefRev[M + N - i - 1];
					if(cur + ofc >= M) {
						tot++; continue www;
					}
				}
			}
			out.println(tot);
		}
		
		void kmp(int[] text, int[] res, int n) {
			int k = 0; res[0] = 0;
			for(int i = 1; i < n; i++) {
				while(k > 0 && text[i] != text[k]) k = res[k - 1];
				if(text[i] == text[k]) k++;
				res[i] = k;
			}
		}
		
		int max(int a, int b) { return a > b ? a : b; }
		
	}
	
	public static void main(String[] args) {
		FastScanner s = new FastScanner(System.in);
		PrintWriter out = new PrintWriter(System.out);
		new Solver().solve(s, out);
		out.close(); s.close();
	}
	
	static class FastScanner {
		private InputStream stream;
		private byte[] buf = new byte[1 << 22];
		private int curChar;
		private int numChars;
		
		public FastScanner(InputStream stream) {
			this.stream = stream;
		}
		
		public FastScanner(File f) throws FileNotFoundException {
			this(new FileInputStream(f));
		}
		
		public FastScanner(String s) {
			this.stream = new ByteArrayInputStream(s.getBytes(StandardCharsets.UTF_8));
		}
		
		void close() {
			try {
				stream.close();
			} catch (IOException e) {
				e.printStackTrace();
			}
		}
		
		int read() {
			if (numChars == -1)
				throw new InputMismatchException();
			if (curChar >= numChars) {
				curChar = 0;
				try {
					numChars = stream.read(buf);
				} catch (IOException e) {
					throw new InputMismatchException();
				}
				if (numChars <= 0)
					return -1;
			}
			return buf[curChar++];
		}
		
		boolean isSpaceChar(int c) {
			return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
		}
		
		boolean isEndline(int c) {
			return c == '\n' || c == '\r' || c == -1;
		}
		
		public int nextInt() {
			return Integer.parseInt(next());
		}
		
		public long nextLong() {
			return Long.parseLong(next());
		}
		
		public double nextDouble() {
			return Double.parseDouble(next());
		}
		
		public String next() {
			int c = read();
			while (isSpaceChar(c))
				c = read();
			StringBuilder res = new StringBuilder();
			do {
				res.appendCodePoint(c);
				c = read();
			} while (!isSpaceChar(c));
			return res.toString();
		}
		
		public String nextLine() {
			int c = read();
			while (isEndline(c))
				c = read();
			StringBuilder res = new StringBuilder();
			do {
				res.appendCodePoint(c);
				c = read();
			} while (!isEndline(c));
			return res.toString();
		}
		
		public int[] nextIntArray(int N) {
			int[] ret = new int[N];
			for (int i = 0; i < N; i++)
				ret[i] = this.nextInt();
			return ret;
		}
		
		public int[][] next2DIntArray(int N, int M) {
			int[][] ret = new int[N][];
			for (int i = 0; i < N; i++)
				ret[i] = this.nextIntArray(M);
			return ret;
		}
		
		public long[] nextLongArray(int N) {
			long[] ret = new long[N];
			for (int i = 0; i < N; i++)
				ret[i] = this.nextLong();
			return ret;
		}
		
		public long[][] next2DLongArray(int N, int M) {
			long[][] ret = new long[N][];
			for (int i = 0; i < N; i++)
				ret[i] = nextLongArray(M);
			return ret;
		}
		
		public double[] nextDoubleArray(int N) {
			double[] ret = new double[N];
			for (int i = 0; i < N; i++)
				ret[i] = this.nextDouble();
			return ret;
		}
		
		public double[][] next2DDoubleArray(int N, int M) {
			double[][] ret = new double[N][];
			for (int i = 0; i < N; i++)
				ret[i] = this.nextDoubleArray(M);
			return ret;
		}
		
	}
	
}