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;
use std::cmp;
use std::collections::BinaryHeap;
use std::collections::HashMap;
use std::collections::HashSet;
use std::f64::consts::PI;

fn read_line() -> String {
    let mut buf = String::new();
    io::stdin().read_line(&mut buf).ok();
    buf
}

const MOD: i64 = 1_000_000_007;

fn extgcd(a: i64, b: i64, x: &mut i64, y: &mut i64) -> i64 {
    let mut d = a;
    if b != 0 {
        d = extgcd(b, a % b, y, x);
        *y -= (a / b) * (*x);
    } else {
        *x = 1; *y = 0;
    }
    d
}

fn mod_inverse(a: i64, m: i64) -> i64 {
    let mut x = 0;
    let mut y = 0;
    extgcd(a, m, &mut x, &mut y);
    (m + x % m) % m
}

fn mod_fact(n: i64, p: i64, e: &mut i64, fact: &Vec<i64>) -> i64 {
    *e = 0;
    if n == 0 {
        return 1;
    }

    let res = mod_fact(n/p, p, e, fact);
    *e += n/p;
    if (n/p %2 != 0) {
        return res * (p - fact[(n%p) as usize]) % p;
    }
    res * fact[(n%p) as usize] % p
}

fn mod_comb(n: i64, k: i64, memo: &Vec<i64>) -> i64 {
    let mut ans = 1;
    for i in 0..k {
        ans *= (n-i);
        ans %= MOD;
    }
    let mut div = memo[k as usize];
    ans * mod_inverse(div, MOD) % MOD
}


fn main() {
    let line: Vec<usize> = read_line().split_whitespace().map(|s| s.parse().unwrap()).collect();
    let n = line[0];
    let a = line[1];
    let b = line[2];
    let mut memo = vec![0i64; cmp::max(a,b)+10];
    memo[0] = 1;
    memo[1] = 1;
    for i in 2..memo.len() {
        memo[i] = memo[i-1] * (i as i64) % MOD;
    }
    let mut ans = 1;
    let mut tmp = 2;
    for i in 0..50 {
        if (n as i64) & (1 << i) > 0 {
            ans = ans * tmp % MOD;
        }
        tmp = tmp * tmp % MOD;  
    }
    ans += MOD-1;
    ans %= MOD;
    ans += MOD - mod_comb(n as i64, a as i64, &memo);
    ans %= MOD;
    ans += MOD - mod_comb(n as i64, b as i64, &memo);
    ans %= MOD;
    println!("{}", ans);
}