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)

val MOD = 1000000007L

fun main(args : Array<String>) {
    val (n, a, b) = readLine()!!.split(" ").map { it.toLong() }

    val all = modPow(2L, n, MOD) - 1L
    val nCa = modComb(n, a, MOD)
    val nCb = modComb(n, b, MOD)

    val tmp = (nCa + nCb)%MOD

    if (all < tmp) {
        println((all-tmp)%MOD+MOD)
    } else {
        println(all-tmp)
    }
}

fun modPow(base: Long, p: Long, m: Long): Long {
    return when {
        p == 0L -> 1L
        p == 1L -> base
        p % 2L == 0L -> modPow(base, p / 2L, m).let { it*it%m }
        else -> base * modPow(base, p - 1L, m) % m
    }
}

fun modComb(n: Long, k: Long, m: Long): Long {
    if (n < k) return 0L
    val kk = Math.min(k, n-k)

    var n1 = 1L
    for (i in 1..kk) {
        n1 *= (n-i+1)
        n1 %= m
    }

    var n2 = 1L
    for (i in 1..kk) {
        n2 *= i
        n2 %= m
    }

    return n1 * modInv(n2, m) % m
}

fun modInv(n: Long, m: Long): Long {
    return modPow(n, m-2, m)
}