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: Wrong Answer)

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

#(a*x)をmで割った余りが1になるようなxを求める
#aで割ったときの余りはa^(-1)をかけた余りと同じ
#逆元a^(-1)を求める
#aとmが互いに素であることが条件

def gyakugen( a, m ):
  b = m
  x, y = 1, 0
  while b>0:
    q = a // b
    a, b = b, a-q*b
    x, y = y, x-q*y
  return x if x>0 else x+m

#nPrをmで割った余りを求める(ただしn<m、mは素数)
import sys
sys.setrecursionlimit(10**9)#再帰の回数の上限を設定

def perm_remainder(n,r,m):
    if r>=1:
        return (perm_remainder(n-1,r-1,m)*n)%m
    else:
        return 1

#nCrをmで割った余りを求める(ただしn<m,mは素数)

def comb_remainder(n,r,m):
    r=min(r,n-r)
    return (perm_remainder(n,r,m)*gyakugen(perm_remainder(r,r,m),m))%m

#(a**n)を素数mで割った余りを求める
#当然aはmで割り切れない整数

def power_remainder(a,n,m):
    a=a%m
    bi=str(format(n,"n"))#2進表現に
    res=1
    for i in range(len(bi)):
        res=(res*res) %m
        if bi[i]=="1":
            res=(res*a) %m
    return res

#2^n-nCa-nCb-1を計算
print(power_remainder(2,n,m)-1-comb_remainder(n,a,m)-comb_remainder(n,b,m))