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)

use std::io;
use std::str::FromStr;

static MOD: u64 = 1_000_000_007;

fn input_tuple() -> (u64, u64, u64) {
    let mut line: String = String::new();
    io::stdin().read_line(&mut line);
    let vec: Vec<u64> = line.split_whitespace().map(|s| s.parse().unwrap()).collect();
    (vec[0], vec[1], vec[2])
}

fn binary_power(x: u64, n: u64) -> u64 {
    if n == 0 {
      return 1;
    }
    if n % 2 == 0 {
      return (binary_power(x, n / 2) % MOD).pow(2) % MOD;
    } else {
      return binary_power(x, n - 1) % MOD * x % MOD;
    }
}
  
fn inverse_mod(n: u64) -> u64 {
    binary_power(n, MOD-2)
}  

fn combination(n: u64, mut m: u64) -> u64 {
    if n < m {
        return 0;
    }
    
    if m > n - m {
        m = n - m
    }

    let mut ret = 1u64;
    for i in 0..m {
        ret %= MOD;
        ret *= n - i;        
    }
    ret %= MOD;

    let mut div = 1u64;
    for i in 1..(m+1) {
        div %= MOD;
        div *= i;
    }
    div %= MOD;

    // println!("{}, {}", ret, div);
    ret * inverse_mod(div) % MOD
}



fn main() {
    let (n, a, b) = input_tuple();
    let ans = binary_power(2, n) + MOD - combination(n, a) + MOD - combination(n, b);

    println!("{}", (ans - 1) % MOD);
}