Problem:
Score : 400 points

Problem Statement

Akari has n kinds of flowers, one of each kind.

She is going to choose one or more of these flowers to make a bouquet.

However, she hates two numbers a and b, so the number of flowers in the bouquet cannot be a or b.

How many different bouquets are there that Akari can make?

Find the count modulo (10^9 + 7).

Here, two bouquets are considered different when there is a flower that is used in one of the bouquets but not in the other bouquet.

Constraints

All values in input are integers.

2 \leq n \leq 10^9

1 \leq a < b \leq \textrm{min}(n, 2 \times 10^5)

Input

Input is given from Standard Input in the following format:

n a b

Output

Print the number of bouquets that Akari can make, modulo (10^9 + 7). (If there are no such bouquets, print 0.)

Sample Input 1

4 1 3

Sample Output 1

7

In this case, Akari can choose 2 or 4 flowers to make the bouquet.

There are 6 ways to choose 2 out of the 4 flowers, and 1 way to choose 4, so there are a total of 7 different bouquets that Akari can make.

Sample Input 2

1000000000 141421 173205

Sample Output 2

34076506

Print the count modulo (10^9 + 7).

Solution in Ruby (2.7.1) (judge verdict: Wrong Answer)

$MOD = 10 ** 9 + 7

def modpow(_n, _pp)
    r = 1
    n = _n
    pp = _pp
    while pp > 0
        r = (r*n) % $MOD if pp&1 == 1
        n = (n*n) % $MOD
        pp = pp >> 1
    end
    return r
end

def fermatComb(n, k)
    fermatComb(n, n - k) if k > n - k
    denominator = 1 # 分母
    numerator = 1 # 分子
    k.times do |i|
        denominator *= (n - i)
        denominator %= $MOD
        numerator *= (i + 1)
        numerator %= $MOD
    end

    return denominator * modpow(numerator, $MOD - 2) % $MOD
end

n, a, b = gets.chomp.split.map(&:to_i)
ans = modpow(2, n) - 1
ans -= (fermatComb(n, a) + fermatComb(n, b)) % $MOD

puts ans