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)

use std::io::*;
const MOD: u64 = 1_000_000_007;

fn mod_pow(mut x: u64, mut e: u64, m: u64) -> u64 {
    let mut res = 1;
    while e > 0 {
        if e & 1 == 1 {
            res = res * x % m;
        }
        x = x * x % m;
        e >>= 1;
    }
    res
}

fn main() {
    let mut s: String = String::new();
    std::io::stdin().read_to_string(&mut s).ok();
    let mut itr = s.trim().split_whitespace();

    let n: u64 = itr.next().unwrap().parse().unwrap();
    let a: u64 = itr.next().unwrap().parse().unwrap();
    let b: u64 = itr.next().unwrap().parse().unwrap();

    let mut ans = mod_pow(2, n, MOD) - 1;

    let mut nca = 1;
    let mut ncb = 1;
    for i in 0..a {
        nca = nca * (n - i) % MOD * mod_pow(a - i, MOD - 2, MOD) % MOD;
    }
    for i in 0..b {
        ncb = ncb * (n - i) % MOD * mod_pow(b - i, MOD - 2, MOD) % MOD;
    }
    ans = (ans + MOD - nca) % MOD;
    ans = (ans + MOD - ncb) % MOD;
    println!("{}", ans);
}