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::cmp;
use std::collections::HashMap;
use std::collections::VecDeque;
use std::usize::MAX;

macro_rules! input {
    (source = $s:expr, $($r:tt)*) => {
        let mut iter = $s.split_whitespace();
        let mut next = || { iter.next().unwrap() };
        input_inner!{next, $($r)*}
    };
    ($($r:tt)*) => {
        let stdin = std::io::stdin();
        let mut bytes = std::io::Read::bytes(std::io::BufReader::new(stdin.lock()));
        let mut next = move || -> String{
            bytes
                .by_ref()
                .map(|r|r.unwrap() as char)
                .skip_while(|c|c.is_whitespace())
                .take_while(|c|!c.is_whitespace())
                .collect()
        };
        input_inner!{next, $($r)*}
    };
  }

macro_rules! input_inner {
    ($next:expr) => {};
    ($next:expr, ) => {};
    ($next:expr, $var:ident : $t:tt $($r:tt)*) => {
        let $var = read_value!($next, $t);
        input_inner!{$next $($r)*}
    };
  }

macro_rules! read_value {
    ($next:expr, ( $($t:tt),* )) => {
        ( $(read_value!($next, $t)),* )
    };
    ($next:expr, [ $t:tt ; $len:expr ]) => {
        (0..$len).map(|_| read_value!($next, $t)).collect::<Vec<_>>()
    };
    ($next:expr, chars) => {
        read_value!($next, String).chars().collect::<Vec<char>>()
    };
    ($next:expr, usize1) => {
        read_value!($next, usize) - 1
    };
    // 3
    // 1 2
    // 1 3
    // 1 4
    // のようなタイプ
    ($next:expr, [ $t:tt ]) => {{
        let len = read_value!($next, usize);
        (0..len).map(|_| read_value!($next, $t)).collect::<Vec<_>>()
    }};
    ($next:expr, $t:ty) => {
        $next().parse::<$t>().expect("Parse error")
    };
  }

// 最大公約数
fn gcd(a: u64, b: u64) -> u64 {
    let mut aa: u64 = if a > b { a } else { b };
    let mut bb: u64 = if a > b { b } else { a };
    while bb != 0 {
        let tmp = bb;
        bb = aa % tmp;
        aa = tmp;
    }
    return aa;
}

// 最小公倍数
fn lcm(a: u64, b: u64) -> u64 {
    a / gcd(a, b) * b
}

// DFS,BFS
fn BFS() {
    let mut searchResult = vec![vec![MAX; 100]; 100];
    let mut q: VecDeque<(usize, usize)> = VecDeque::new();
    q.push_front((0, 0));
    searchResult[0][0] = 0;

    let moves = [(1, 0), (-1, 0), (0, 1), (0, -1)];

    while !q.is_empty() {
        let pos = q.pop_back().unwrap(); // これがpop_frontならstackになって深さ優先になる
        for m in moves.iter() {
            let dx: i64 = pos.0 as i64 + m.0;
            let dy: i64 = pos.1 as i64 + m.1;
            if dx >= 0
                && dx < 100
                && dy >= 0
                && dy < 100
                && searchResult[dx as usize][dy as usize] == MAX
            {
                q.push_front((dx as usize, dy as usize));
                searchResult[dx as usize][dy as usize] = searchResult[pos.0][pos.1] + 1;
            }
        }
    }
}

// 文字列操作

const LARGE_PRIME: u64 = 1_000_000_007;

fn power_mod(base: u64, exp: u64, modulo: u64) -> u64 {
    if exp == 0 {
        return 1;
    } else if exp & 1 == 0 {
        return (power_mod(base, exp / 2, modulo) % modulo).pow(2) % modulo;
    } else {
        return power_mod(base, exp - 1, modulo) % modulo * base % modulo;
    }
}

fn inverse_mod(element: u64, modulo: u64) -> u64 {
    // フェルマーの小定理からpが素数なら a^(p-1) = 1
    // よってa*a^(p-2) = 1 より a^(p-2)がaの逆元

    power_mod(element, modulo - 2, modulo) % modulo
}

fn combination_mod(m: u64, n: u64, modulo: u64) -> u64 {
    // m C n = m! / ( n! * (m - n)! )
    //       = m*(m-1)*(m-2)*...*(m-n+1)/n!
    //       = m*(m-1)*(m-2)*...*(m-n+1) * (n!)^-1

    let mut numerator = 1;
    for i in 0..n {
        numerator *= m - i;
        numerator %= modulo;
    }

    // (m-n)!
    let mut element = 1;
    for j in 1..(n + 1) {
        element *= j;
        element %= modulo;
    }

    numerator * inverse_mod(element, modulo) % modulo
}

fn topological_sort(n: usize, edges: Vec<(usize, usize)>) -> Vec<usize> {
    let mut result = vec![];
    let mut isolated = VecDeque::new();

    // to[x] = [a,b] x -> a, x -> b が存在
    let mut to = vec![vec![]; n + 1];

    // deg[x] = a -> x, b -> xのような辺の数
    let mut deg = vec![0; n + 1];
    for e in edges {
        to[e.0].push(e.1);
        deg[e.1] += 1;
    }

    for i in 1..n + 1 {
        if deg[i] == 0 {
            isolated.push_front(i);
        }
    }

    while isolated.len() != 0 {
        let next = isolated.pop_back().unwrap();
        result.push(next);
        for i in &to[next] {
            deg[*i] -= 1;
            if deg[*i] == 0 {
                isolated.push_front(*i);
            }
        }
    }

    result
}

fn solve() {
    input! {
        n: usize,
        a: usize,
        b: usize
    }

    let mut ans = 0;

    let a_comb = combination_mod(n as u64, a as u64, LARGE_PRIME);
    let b_comb = combination_mod(n as u64, b as u64, LARGE_PRIME);

    let sumnck = power_mod(2, n as u64, LARGE_PRIME);

    ans = sumnck - 1;
    if ans < a_comb {
        ans += LARGE_PRIME;
    }
    ans -= a_comb;
    if ans < b_comb {
        ans += LARGE_PRIME;
    }
    ans -= b_comb;
    println!("{}", ans);
}

fn main() {
    let thd = std::thread::Builder::new().stack_size(104_857_600);
    thd.spawn(|| solve()).unwrap().join().unwrap();
}