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: Time Limit Exceeded)

# coding: utf-8

import sys
import math
import collections
import itertools
from inspect import currentframe
INF = 10 ** 10
MOD = 10 ** 9 + 7
def input() : return sys.stdin.readline().strip()
def gcd(x, y) : return y if x % y == 0 else gcd(y, x % y)
def lcm(x, y) : return (x * y) // gcd(x, y)
def I() : return int(input())
def MI() : return map(int, input().split())
def LI() : return [int(x) for x in input().split()]
def RI(N) : return [int(input()) for _ in range(N)]
def LRI(N) : return [[int(x) for x in input().split()] for _ in range(N)]
def chkprint(*args) : names = {id(v):k for k,v in currentframe().f_back.f_locals.items()}; print(', '.join(names.get(id(arg),'???')+' = '+repr(arg) for arg in args))

def ncr(n, r):
    a = 1
    b = 1
    for i in range(r):
        a *= (n - i) % MOD
        b *= (r - i) % MOD
    
    return a // b % MOD

N, A, B = MI()

s = pow(2, N, MOD) - 1



a = ncr(N, A)
b = ncr(N, B)

print((s - a - b)  % MOD)