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 Kotlin (1.0.0) (judge verdict: Accepted)

fun main(args: Array<String>) = abc156d()

fun abc156d() {
    val (n, a, b) = readLine()!!.split(' ').map { it.toInt() }

    var answer = modPow(2, n.toLong())
    answer -= 1
    answer = (answer + MOD) % MOD
    answer -= comb(n, a)
    answer = (answer + MOD) % MOD
    answer -= comb(n, b)
    answer = (answer + MOD) % MOD

    println(answer)
}

private const val MOD = (1e9 + 7).toLong()

private fun modPow(n: Long, p: Long): Long {
    var x = n
    var y = p
    var result = 1L
    x %= MOD
    while (y > 0) {
        if (y % 2 == 1L)
            result = (result * x) % MOD
        y = y shr 1
        x = (x * x) % MOD
    }
    return result
}

private fun comb(n: Int, k: Int): Long {
    if (n < k) return 0
    val min = Math.min(k, n - k)
    var value = 1L
    for (i in 1L..min) {
        value *= n - i + 1
        value %= MOD
        value *= modPow(i, MOD - 2)
        value %= MOD
    }
    return value
}