Petr stands in line of n people, but he doesn't know exactly which position he occupies. He can say that there are no less than a people standing in front of him and no more than b people standing behind him. Find the number of different positions Petr can occupy.

Input

The only line contains three integers n, a and b (0 ≤ a, b < n ≤ 100).

Output

Print the single number — the number of the sought positions.

Examples

Input

3 1 1


Output

2


Input

5 2 3


Output

3

Note

The possible positions in the first sample are: 2 and 3 (if we number the positions starting with 1).

In the second sample they are 3, 4 and 5.
//package CF;  
import java.io.*;
import java.util.*;
public class B{

    public static void main(String[] args) throws Exception {
	Scanner bf = new Scanner(System.in);
	PrintWriter out = new PrintWriter(System.out);
	int n = bf.nextInt();
	int a = bf.nextInt();
	int b = bf.nextInt();
	
	out.println(1+n-Math.max(a+1, n-b));
	out.flush();
	out.close();
    }

    static class Scanner 
    {
	StringTokenizer st;
	BufferedReader br;

	public Scanner(InputStream s){	br = new BufferedReader(new InputStreamReader(s));}

	public String next() throws IOException 
	{
	    while (st == null || !st.hasMoreTokens()) 
		st = new StringTokenizer(br.readLine());
	    return st.nextToken();
	}

	public int nextInt() throws IOException {return Integer.parseInt(next());}

	public long nextLong() throws IOException {return Long.parseLong(next());}

	public String nextLine() throws IOException {return br.readLine();}

	public double nextDouble() throws IOException
	{
	    String x = next();
	    StringBuilder sb = new StringBuilder("0");
	    double res = 0, f = 1;
	    boolean dec = false, neg = false;
	    int start = 0;
	    if(x.charAt(0) == '-')
	    {
		neg = true;
		start++;
	    }
	    for(int i = start; i < x.length(); i++)
		if(x.charAt(i) == '.')
		{
		    res = Long.parseLong(sb.toString());
		    sb = new StringBuilder("0");
		    dec = true;
		}
		else
		{
		    sb.append(x.charAt(i));
		    if(dec)
			f *= 10;
		}
	    res += Long.parseLong(sb.toString()) / f;
	    return res * (neg?-1:1);
	}

	public boolean ready() throws IOException {return br.ready();}
    }

}