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: Wrong Answer)

use proconio::{input, fastout};

const MOD: usize = 1_000_000_007;

#[fastout]
fn main() {
    input! {
        n: usize, a: usize, b: usize,
    }

    let mut ans = 1;
    let mut last_binom = 1;
    let aa = if a > n/2 { n-a } else { a };
    let bb = if b > n/2 { n-b } else { b };
    for i in 1..(n+1)/2 {
        let c = last_binom * (n+1-i) / i % MOD;
        let mut d = 2;
        if aa == i { d -= 1};
        if bb == i { d -= 1};
        ans = (ans + d*c) % MOD;
        last_binom = c;
    }
    if n%2 == 0 && n/2 != aa && n/2 != bb {
        ans = (ans + last_binom * (n + 2) / n) % MOD;
    }

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