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 PyPy3 (2.4.0) (judge verdict: Accepted)

mod=10**9+7
n,a,b=map(int,input().split())
ans=pow(2,n,mod)-1 #n種類の花を使って花束を作る総数は2^n-1通り
comb1=1 #nCaの計算
for i in range(n-a+1,n+1): #n*(n-1)*…*(n-a+1)を計算
 comb1*=i
 comb1%=mod
for i in range(1,a+1): #1/a!=1/(1*2*…*a)=1^(p-2)*2^(p-2)*…*a^(p-2) (p=10**9+7)を計算
 comb1*=pow(i,mod-2,mod)
 comb1%=mod
comb2=1
for i in range(n-b+1,n+1): #n*(n-1)*…*(n-b+1)を計算
 comb2*=i
 comb2%=mod
for i in range(1,b+1): #1/b!=1/(1*2*…*b)=1^(p-2)*2^(p-2)*…*b^(p-2) (p=10**9+7)を計算
 comb2*=pow(i,mod-2,mod)
 comb2%=mod
ans-=(comb1+comb2) #nCa、nCbを総数から引く
ans%=mod
print(ans)