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 PHP7 (7.0.15) (judge verdict: Accepted)

<?php
fscanf(STDIN, '%d %d %d', $n, $a, $b);
define('MOD', 10 ** 9 + 7);

$ans = modSub(modSub(modSub(modPow(2, $n), 1), nCr($n, $a)), nCr($n, $b));
echo $ans . PHP_EOL;

// 足し算
function modAdd($x, $y)
{
    return ($x + $y) % MOD;
}

// 引き算
function modSub($x, $y)
{
    return ($x + MOD - $y) % MOD;
}

// 掛け算
function modMul($x, $y)
{
    return ($x * $y) % MOD;
}

// 割り算
function modDiv($x, $y)
{
    return $x * modPow($y, MOD - 2) % MOD;
}

// 累乗（繰り返し二乗法）
function modPow($n, $x)
{
    if ($x === 0) return 1;
    $res = (modPow($n, $x >> 1) ** 2) % MOD;
    if ($x % 2 === 1) $res = ($res * $n) % MOD;
    return $res;
}

// 階乗
function modFac($n)
{
    if ($n === 0) return 1;
    return $n * modFac($n - 1) % MOD;
}

// 順列
function nPr($n, $r)
{
    if ($r === 0) return 1;
    return nPr($n, $r - 1) * ($n - $r + 1) % MOD;
}

// 組み合わせ
function nCr($n, $r)
{
    $r = min($r, $n - $r);
    if ($r === 0) return 1;
    if ($r === 1) return $n;
    return modDiv(nPr($n, $r), modFac($r));
}