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)

import java.lang.Math

fun main(args: Array<String>) {
    val mod = 1000000007L

    val (n, a, b) = readInputLine().split(" ").map { it.toInt() }

    val inverse = LongArray(Math.min(n, 200000) + 1)
    val factorialInverse = LongArray(Math.min(n, 200000) + 1)
    inverse[0] = 1L
    inverse[1] = 1L
    factorialInverse[0] = 1L
    factorialInverse[1] = 1L
    for (i in 2 until Math.min(n, 200000) + 1) {
        inverse[i] = (mod - inverse[mod.toInt() % i] * (mod / i.toLong()) % mod)
        factorialInverse[i] = (factorialInverse[i - 1] * inverse[i] % mod)
    }

    // 2^n
    val pow = sortedMapOf(Pair(0, 1L), Pair(1, 2L))
    var powNum = 2
    while (powNum <= n) {
        pow[powNum] = (pow[powNum / 2]!! * pow[powNum / 2]!!) % mod
        powNum *= 2
    }
    pow[powNum] = (pow[powNum / 2]!! * pow[powNum / 2]!!) % mod

    var ans = 1L

    var nTmp = n
    powNum = 1

    while (nTmp != 0) {
        if (nTmp % 2 == 1) {
            ans *= pow[powNum]!!
            ans %= mod
        }
        powNum *= 2
        nTmp /= 2
    }

    ans = (ans + mod - 1L) % mod

    var subTmp = 1L

    for (i in (n - a) + 1..n) {
        subTmp *= i.toLong()
        subTmp %= mod
    }
    subTmp *= factorialInverse[a]
    subTmp %= mod

    ans = (ans + mod - subTmp) % mod

    subTmp = 1L

    for (i in (n - b) + 1..n) {
        subTmp *= i.toLong()
        subTmp %= mod
    }
    subTmp *= factorialInverse[b]
    subTmp %= mod

    ans = (ans + mod - subTmp) % mod

    println(ans)
}

fun readInputLine(): String {
    return readLine()!!
}