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 = map(int, input().split())

p = 10 ** 9 + 7
N = 10 ** 5 *5  # N は必要分だけ用意する
fact = [1, 1]  # fact[n] = (n! mod p)
factinv = [1, 1]  # factinv[n] = ((n!)^(-1) mod p)
inv = [0, 1]  # factinv 計算用

for i in range(2, N + 1):
    fact.append((fact[-1] * i) % p)
    inv.append((-inv[p % i] * (p // i)) % p)
    factinv.append((factinv[-1] * inv[-1]) % p)


# def f(x):
#     if x==1:
#         return 2
#     elif x==2:
#         return 4
#     elif x%2==0:
#         return f(x//2)**2
#     else:
#         return f(x//2)**2*2

di131072 = 1
for i in range(131072):
    di131072 *= 2
    di131072 %= p

q = n//131072
r = n%131072
ans = 1
for i in range(q):
    ans *= di131072
    ans %= p
for i in range(r):
    ans *= 2
    ans %= p


fact2 = []
fact2num = max(a,b)+2
now = 1
for i in range(fact2num):
    now *= (n-i)
    now %= p
    fact2.append(now)

def ncr(n,r):
    return (fact2[r-1]*factinv[r])%p


print((ans-1-ncr(n,a)-ncr(n,b))%p)