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.
The only line contains three integers *n*, *a* and *b* (0 ≤ *a*, *b* &lt; *n* ≤ 100).
3 1 1
2
5 2 3
3
5 4 0
1
6 5 5
1
9 4 3
4
11 4 6
7
13 8 7
5
14 5 5
6
16 6 9
10
20 13 17
7
22 4 8
9
23 8 14
15
26 18 22
8
28 6 1
2
29 5 23
24
32 27 15
5
33 11 5
6
37 21 15
16
39 34 33
5
41 27 11
12
42 25 16
17
45 7 43
38
47 16 17
18
49 11 37
38
51 38 39
13
52 29 7
8
56 43 12
13
58 57 28
1
59 12 39
40
62 9 52
53
63 29 44
34
65 30 22
23
66 27 38
39
71 33 53
38
73 14 12
13
73 37 35
36
76 69 44
7
79 25 20
21
81 60 20
21
81 79 14
2
84 0 42
43
88 79 8
9
90 76 59
14
92 2 22
23
94 5 88
89
94 62 48
32
96 22 72
73
100 11 88
89
100 81 91
19
1 0 0
1
import math, heapq
from sys import stdin	
from collections import Counter, defaultdict, deque, namedtuple, OrderedDict
from bisect import bisect_right, bisect_left
from typing import List, DefaultDict
from itertools import permutations, combinations
from copy import deepcopy

 
def sieve_of_eratosthenes(n):
    is_prime = [True] * (n+1)
    is_prime[0] = is_prime[1] = False
    for p in range(2, int(n**0.5)+1):
        if is_prime[p]:
            for i in range(p*p, n+1, p):
                is_prime[i] = False
                
    for i in range(2, n+1):
        if is_prime[i]:
            print(i, end=' ')


def isPowerOf2(n):

    def log2(x):

        if x == 0:
            return False

        return (math.log10(x)/math.log10(2))


    return math.ceil(log2(n)) == math.floor(log2(n))


def binPowRec(a,b):

    if b == 0:
        return 1
    
    res = binPowRec(a, b//2)
    # If b is odd
    # If b % 2 == 1
    if b % 2:
        return res * res * a
    else:
        return res * res


# pow(a,b) in O(log n)
def binPow(a, b):

    res = 1

    while b > 0:
        # b << 1 = b * 2
        # b >> 1 = b/2
        # If b & 1 = b % 2 
        # If b % 2 == 1
        if b & 1:
            res = res * a

        a = a * a

        # b >> 1  = b//2
        # b >> 2  = b//4
        # b >> 3  = b//8
        # b >> 4  = b//16
        b >>= 1

    return res

 
def readarray(typ):
    return list(map(typ, stdin.readline().split()))


def readint():
    return int(input())


n, a, b = readarray(int)


ans = n-a

while b > 0 and a > b:
    a -= 1
    ans -= 1

print(ans)