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)

import sys
import math
import fractions
import itertools
from collections import deque
import copy
import bisect

MOD = 10 ** 9 + 7
INF = 10 ** 18 + 7
input = lambda: sys.stdin.readline().strip()


def inv_mod(a, N):
    """
    ユークリッドの互除法より
    aの逆元 ax = 1 (mod N)を満たすxを返す。
    O(log a)
    """

    def extgcd(a, b):
        """
        Extended Great Common Divisor
        ax + by = gcd(a, b) = dを満たすx, y, dを返す。
        """
        d = fractions.gcd(a, b)
        if b == 0:
            return 1, 0, a
        else:
            q, r = divmod(a, b)
            x, y, d = extgcd(b, r)
            s, t = y, x - q * y
            return s, t, d

    return extgcd(a, N)[0]


def bpm(x, y, N):
    """
     Bisection power method : 二分累乗法
     O(log y)でx^(y) mod N を求める。
    """
    ans = 1
    x %= MOD
    digit = int((math.log(y, 2) // 1 + 1))  # yの２進数表示の桁数
    for i in range(digit):
        if (y >> i) & 1 == 1:
            ans *= x
            ans %= MOD
        x **= 2
        x %= MOD
    return ans


n, a, b = map(int, input().split())

nCa = 1
nCb = 1

inv_list = []

for i in range(max(a, b)):
    inv_list.append(inv_mod(i + 1, MOD))

for i in range(a):
    nCa *= inv_list[i]
    nCa %= MOD
    nCa *= n - i
    nCa %= MOD

for i in range(b):
    nCb *= inv_list[i]
    nCb %= MOD
    nCb *= n - i
    nCb %= MOD

print((bpm(2, n, MOD) - 1 - nCa - nCb) % MOD)