Leonid works for a small and promising start-up that works on decoding the human genome. His duties include solving complex problems of finding certain patterns in long strings consisting of letters 'A', 'T', 'G' and 'C'.

Let's consider the following scenario. There is a fragment of a human DNA chain, recorded as a string S. To analyze the fragment, you need to find all occurrences of string T in a string S. However, the matter is complicated by the fact that the original chain fragment could contain minor mutations, which, however, complicate the task of finding a fragment. Leonid proposed the following approach to solve this problem.

Let's write down integer k ≥ 0 — the error threshold. We will say that string T occurs in string S on position i (1 ≤ i ≤ |S| - |T| + 1), if after putting string T along with this position, each character of string T corresponds to the some character of the same value in string S at the distance of at most k. More formally, for any j (1 ≤ j ≤ |T|) there must exist such p (1 ≤ p ≤ |S|), that |(i + j - 1) - p| ≤ k and S[p] = T[j].

For example, corresponding to the given definition, string "ACAT" occurs in string "AGCAATTCAT" in positions 2, 3 and 6.

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Note that at k = 0 the given definition transforms to a simple definition of the occurrence of a string in a string.

Help Leonid by calculating in how many positions the given string T occurs in the given string S with the given error threshold.

Input

The first line contains three integers |S|, |T|, k (1 ≤ |T| ≤ |S| ≤ 200 000, 0 ≤ k ≤ 200 000) — the lengths of strings S and T and the error threshold.

The second line contains string S.

The third line contains string T.

Both strings consist only of uppercase letters 'A', 'T', 'G' and 'C'.

Output

Print a single number — the number of occurrences of T in S with the error threshold k by the given definition.

Examples

Input

10 4 1
AGCAATTCAT
ACAT


Output

3

Note

If you happen to know about the structure of the human genome a little more than the author of the problem, and you are not impressed with Leonid's original approach, do not take everything described above seriously.
/*
 * Code Author: Akshay Miterani
 * DA-IICT
 */
import java.io.*;
import java.math.*;
import java.util.*;

public class Main {

	InputReader in;
	PrintWriter out;
	Main() throws IOException {
		in = new InputReader(System.in);
		out = new PrintWriter(System.out);
		//in = new InputReader(new FileInputStream("check.txt"));
		//out = new PrintWriter(new File("check.txt"));
		solve();
		out.close();
	}

	public static void main(String args[]) {
		new Thread(null, new Runnable() {
			public void run() {
				try {
					new Main();
				} catch (Exception e) {
					e.printStackTrace();
				}
			}
		}, "1", 1 << 26).start();
	}

	//--------------------My Code Starts Here----------------------


	int mod=(int)1e9+7;
	public void solve() throws IOException{
		int n=in.nextInt();
		int m=in.nextInt();
		int k=in.nextInt();
		String s=in.next();
		String t=in.next();
		String AGCT="AGCT";
		BitSet b[]=new BitSet[4];
		for(int i=0;i<4;i++){
			b[i]=new BitSet(n);
		}
		int c[][]=new int[n+1][4];
		for(int i=0;i<n;i++){
			int ind=AGCT.indexOf(s.charAt(i));
			c[Math.max(0, i-k)][ind]++;
			c[Math.min(n, i+k+1)][ind]--;
		}
		for(int j=0;j<4;j++){
			for(int i=0;i<n;i++){
				if(c[i][j]>0){
					b[j].set(i);
				}
				c[i+1][j]+=c[i][j];
			}
		}
		BitSet res=new BitSet(n);
		res.set(0, n, true);
		for(int i=0;i<m;i++){
			int ind=AGCT.indexOf(t.charAt(i));
			res.and(b[ind].get(i, n));
		}
		out.println(res.cardinality());
	}
	//---------------------My Code Ends Here------------------------
	
	static long modulo(long a,long b,long c) {
		long x=1;
		long y=a;
		while(b > 0){
			if(b%2 == 1){
				x=(x*y)%c;
			}
			y = (y*y)%c; // squaring the base
			b /= 2;
		}
		return  x%c;
	}
	static class Pair implements Comparable<Pair>{
		int x;
		int y;
		Pair(int xx,int yy){
			x=xx;
			y=yy;
		}
		public String toString() {
			return "[x = "+x+","+y+"]";
		}
		@Override
		public int compareTo(Pair o) {
			return -Long.compare(this.y-this.x, o.y-o.x);
		}
	}
	public static void debug(Object... o) {
		System.out.println(Arrays.deepToString(o));
	}

	static final class InputReader{
		private final InputStream stream;
		private final byte[] buf=new byte[1024];
		private int curChar;
		private int numChars;
		public InputReader(InputStream stream){this.stream=stream;}
		private int read()throws IOException{
			if(curChar>=numChars){
				curChar=0;
				numChars=stream.read(buf);
				if(numChars<=0)
					return -1;
			}
			return buf[curChar++];
		}
		public final int nextInt()throws IOException{return (int)nextLong();}
		public final long nextLong()throws IOException{
			int c=read();
			while(isSpaceChar(c)){
				c=read();
				if(c==-1) throw new IOException();
			}
			boolean negative=false;
			if(c=='-'){
				negative=true;
				c=read();
			}
			long res=0;
			do{
				if(c<'0'||c>'9')throw new InputMismatchException();
				res*=10;
				res+=(c-'0');
				c=read();
			}while(!isSpaceChar(c));
			return negative?(-res):(res);
		}
		public final int[] readIntBrray(int size)throws IOException{
			int[] arr=new int[size];
			for(int i=0;i<size;i++)arr[i]=nextInt();
			return arr;
		}
		public final String next()throws IOException{
			int c=read();
			while(isSpaceChar(c))c=read();
			StringBuilder res=new StringBuilder();
			do{
				res.append((char)c);
				c=read();
			}while(!isSpaceChar(c));
			return res.toString();
		}
		public final String nextLine()throws IOException{
			int c=read();
			while(isSpaceChar(c))c=read();
			StringBuilder res=new StringBuilder();
			do{
				res.append((char)c);
				c=read();
			}while(c!='\n'&&c!=-1);
			return res.toString();
		}
		private boolean isSpaceChar(int c){
			return c==' '||c=='\n'||c=='\r'||c=='\t'||c==-1;
		}
	}
}