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.
import java.util.Scanner;
public class Codechef{
	static void merge(int arr[], int l, int m, int r) 
	{ 
		int n1 = m - l + 1; 
		int n2 = r - m;
		int L[] = new int [n1]; 
		int R[] = new int [n2];
		for (int i=0; i<n1; ++i) 
			L[i] = arr[l + i]; 
		for (int j=0; j<n2; ++j) 
			R[j] = arr[m + 1+ j]; 
		int i = 0, j = 0;
		int k = l; 
		while (i < n1 && j < n2) 
		{ 
			if (L[i] <= R[j]) 
			{ 
				arr[k] = L[i]; 
				i++; 
			} 
			else
			{ 
				arr[k] = R[j]; 
				j++; 
			} 
			k++; 
		} 
		while (i < n1) 
		{ 
			arr[k] = L[i]; 
			i++; 
			k++; 
		} 
		while (j < n2) 
		{ 
			arr[k] = R[j]; 
			j++; 
			k++; 
		} 
	}  
	public static void sort(int arr[], int l, int r) 
	{ 
		if (l < r) 
		{
			int m = (l+r)/2;
			sort(arr, l, m); 
			sort(arr , m+1, r); 
			merge(arr, l, m, r); 
		} 
	} 
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);

		int n = sc.nextInt();
		int a = sc.nextInt();
		int b = sc.nextInt();
		int count=0;
		int i = a;
		i++;
		int j = n-b;
		int limit = Math.max(i, j);
		while(limit<=n) {
			count++;
			limit++;
		}
		System.out.println(count);
	}
}