Snuke has N integers: 1,2,\ldots,N. He will choose K of them and give those to Takahashi.

How many ways are there to choose K consecutive integers?

Constraints

* All values in input are integers.
* 1 \leq K \leq N \leq 50

Input

Input is given from Standard Input in the following format:


N K


Output

Print the answer.

Examples

Input

3 2


Output

2


Input

13 3


Output

11
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.NoSuchElementException;
import java.util.PriorityQueue;

class Main {
	public static void main(String[] args) {
		new Main().run();
	}
	
	void run() {
		FastScanner sc=new FastScanner();
		PrintWriter pw=new PrintWriter(System.out);
		int N=sc.nextInt();
		int K=sc.nextInt();
		pw.println(N-K+1);
		pw.close();
	}
	
	static void tr(Object... objects) {
		System.out.println(Arrays.deepToString(objects));
	}

}


class FastScanner {
    private final InputStream in = System.in;
    private final byte[] buffer = new byte[1024];
    private int ptr = 0;
    private int buflen = 0;
    private boolean hasNextByte() {
        if (ptr < buflen) {
            return true;
        }else{
            ptr = 0;
            try {
                buflen = in.read(buffer);
            } catch (IOException e) {
                e.printStackTrace();
            }
            if (buflen <= 0) {
                return false;
            }
        }
        return true;
    }
    private int readByte() { if (hasNextByte()) return buffer[ptr++]; else return -1;}
    private static boolean isPrintableChar(int c) { return 33 <= c && c <= 126;}
    public boolean hasNext() { while(hasNextByte() && !isPrintableChar(buffer[ptr])) ptr++; return hasNextByte();}
    public String next() {
        if (!hasNext()) throw new NoSuchElementException();
        StringBuilder sb = new StringBuilder();
        int b = readByte();
        while(isPrintableChar(b)) {
            sb.appendCodePoint(b);
            b = readByte();
        }
        return sb.toString();
    }

    public int nextInt() {
    	return Integer.parseInt(next());
    }
    
    public long nextLong() {
    	return Long.parseLong(next());
    }
}