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 JavaScript (node.js v5.12) (judge verdict: Runtime Error)

;('use strict')

const main = arg => {
  const n = BigInt(parseInt(arg.split(' ')[0], 10))
  const a = BigInt(parseInt(arg.split(' ')[1], 10))
  const b = BigInt(parseInt(arg.split(' ')[2], 10))
  const p = BigInt(Math.pow(10, 9) + 7)

  const modpow = (a, exp, mod) => {
    if (exp == 0n) return BigInt(1)
    if (exp == 1n) return BigInt(a % mod)
    if (exp % 2n == 0) {
      const d = BigInt(modpow(a, exp / 2n, mod) % mod)
      return BigInt(d * d) % mod
    }
    return BigInt(a * modpow(a, exp - 1n, mod)) % mod
  }

  let ret = modpow(2n, BigInt(n), p) - 1n

  let _a1 = 1n
  let _a2 = 1n
  let _b1 = 1n
  let _b2 = 1n

  for (let i = 0n; i < a; i++) {
    _a1 = BigInt(_a1 * (n - i)) % p
    _a2 = BigInt(_a2 * (i + 1n)) % p
  }
  let _a = BigInt(_a1 * modpow(_a2, p - 2n, p)) % p

  for (let i = 0n; i < b; i++) {
    _b1 = BigInt(_b1 * (n - i)) % p
    _b2 = BigInt(_b2 * (i + 1n)) % p
  }
  let _b = BigInt(_b1 * modpow(_b2, p - 2n, p)) % p

  ret -= _a
  ret -= _b
  console.log(ret.toString())
}
main(require('fs').readFileSync('/dev/stdin', 'utf-8'))