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 Python (3.4.3) (judge verdict: Accepted)

n, a, b = list(map(int, input().split()))
p = (10 ** 9) + 7

def powmod(x, exp, mod):
    if(exp == 1):
        return x % mod

    if(exp % 2 == 1):
        return (x * powmod(x, exp - 1, mod)) % mod

    x = powmod(x, exp / 2, mod)

    return (x * x) % mod

# nCrを計算するが、nもrも巨大なためフェルマーの小定理を利用する
# x = nPr
# y = r!
# x / y ≡ x * y^-1
# p を法とした時、掛けるたびにmodをとった数同士の割り算はうまくいかない
# フェルマーの小定理から
# y^(p-1) ≡ 1
# y * y^(p-2) ≡ 1
# y^(p-2)がyの逆元と同じ振る舞いをするのでy^(p-2)を掛ける
def combination(n, r, mod):
    x = 1
    y = 1

    for i in range(r):
        x *= n - i # n * (n - 1) * (n - 2) * ... * (n - i + 1)
        x %= mod
        y *= i + 1 # r * (r - 1) * (r - 2) * ... * 2 * 1 を逆さまにかけていく
        y %= mod

    return x * powmod(y, mod - 2, mod) % mod


total = powmod(2, n, p) - 1 #1本も花を選ばないパターンは除外する
comb_a = combination(n, a, p)
comb_b = combination(n, b, p)


print((total - comb_a - comb_b) % p)