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)

// -*- coding:utf-8-unix -*-

use proconio::input;

// n^p mod m (繰り返し二乗法)
fn repeat_square(n: u64, p: u64, m: u64) -> u64 {
    if p == 0 {
        1
    } else if p == 1 {
        n % m
    } else if p % 2 == 0 {
        repeat_square(n, p / 2, m).pow(2) % m
    } else {
        (n * repeat_square(n, p - 1, m)) % m
    }
}

// m が素数であること
fn ncr_mod(n: u64, r: u64, m: u64) -> u64 {
    let mut denominator = n;
    let mut numerator = 1;
    for i in 1..r {
        denominator = (denominator * (n - i)) % m;
        numerator = (numerator * (i + 1)) % m;
    }
    (denominator * repeat_square(numerator, m - 2, m)) % m
}

fn main() {
    input! {
        n: u64,
        a: u64,
        b: u64,
    }

    let divisor = 10_u64.pow(9) + 7;
    let mut ans: u64 = repeat_square(2, n, divisor) - 1;

    ans = (ans + divisor - ncr_mod(n, a, divisor)) % divisor;
    ans = (ans + divisor - ncr_mod(n, b, divisor)) % divisor;

    println!("{}", ans);
}