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 Rust (1.42.0) (judge verdict: Accepted)

fn input_trio<T: std::str::FromStr + Copy> () -> (T, T, T) {
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).unwrap();
    let arr: Vec<_> = s.trim().split_whitespace().flat_map(str::parse).collect();
    (arr[0], arr[1], arr[2])
}

fn pow(a: isize, b: isize, m: isize) -> isize {
    if b == 0 {
        return 1;
    }
    let x = pow(a, b >> 1, m);
    return if b % 2 == 0 {
        (x * x) % m
    } else {
        (((a * x) % m) * x) % m
    }
}

fn factorial(n: isize, m: isize) -> isize {
    let mut v = 1;
    for i in 1..=n {
        v = (v * i) % m;
    }
    v
}

fn factorial_inv(n: isize, m: isize) -> isize {
    pow(factorial(n, m), m - 2, m)
}

fn combination(n: isize, r: isize, m: isize) -> isize {
    if r < 0 || r > n {
        0
    } else {
        // ((factorial(n, m) * factorial_inv(r, m)) % m * factorial_inv(n - r, m)) % m
        let mut prev = 1;
        for i in 0..r {
            prev = ((((n - i) * prev) % m) * pow(i + 1, m - 2, m)) % m;
        }
        prev
    }
}

fn main() {
    let (n, a, b) = input_trio::<isize>();
    let m = 1000000007;
    // let mut result = 0;
    // tle
    // for v in (2..=n).filter(|&r| r != a && r != b).map(|r| combination(n, r, m)) {
    //     result = (result + v) % m;
    // }

    // wa, tle
    // let mut prev = n;
    // for i in 1..n {
    //     prev = ((((n - i) * prev) % m) * pow(i + 1, m - 2, m)) % m;
    //     if i != a && i != b {
    //         result = (result + prev) % m;
    //     }
    // }
    // println!("{}", result);

    println!("{}", (pow(2, n, m) - 1 + m - combination(n, a, m) + m - combination(n, b, m)) % m);
    // println!("{}", (pow(2, n, m) - 1 + m - combination(n, a, m)) % m);
}