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 (1.15.1) (judge verdict: Accepted)

fn main() {
    use std::io::Read;
    let mut input = String::new();
    std::io::stdin().read_to_string(&mut input).unwrap();
    println!("{}", solve(&input));
}

const MODULO: i64 = 1_000_000_007;

fn solve(input: &str) -> i64 {
    let mut iter = input.split_whitespace().map(|w| w.parse().unwrap());
    let n: i64 = iter.next().unwrap();
    let a: i64 = iter.next().unwrap();
    let b: i64 = iter.next().unwrap();
    let mut ans = power(2, n);
    ans = (ans + MODULO - 1) % MODULO;
    ans = (ans + MODULO - binom(n, a)) % MODULO;
    ans = (ans + MODULO - binom(n, b)) % MODULO;
    ans
}

fn power(base: i64, exponent: i64) -> i64 {
    let mut acc = 1;
    let mut base = base;
    let mut exponent = exponent;
    while exponent != 0 {
        if exponent & 1 == 1 {
            acc = acc * base % MODULO;
        }
        base = base * base % MODULO;
        exponent >>= 1;
    }
    acc
}

fn inverse(x: i64) -> i64 {
    power(x, MODULO - 2)
}

fn binom(n: i64, k: i64) -> i64 {
    let mut numer = 1;
    let mut denom = 1;
    for i in 0..k {
        numer = numer * (n - i) % MODULO;
        denom = denom * (k - i) % MODULO;
    }
    numer * inverse(denom) % MODULO
}