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)

# python3.4.2用
import math
import fractions
import bisect
import collections
import itertools
import heapq
import string
import sys
import copy
from decimal import *
from collections import deque
sys.setrecursionlimit(10**7)
MOD = 10**9+7
INF = float('inf') #無限大
def gcd(a,b):return fractions.gcd(a,b) #最大公約数
def lcm(a,b):return (a*b) // fractions.gcd(a,b) #最小公倍数
def iin(): return int(sys.stdin.readline()) #整数読み込み
def ifn(): return float(sys.stdin.readline()) #浮動小数点読み込み
def isn(): return sys.stdin.readline().split() #文字列読み込み
def imn(): return map(int, sys.stdin.readline().split()) #整数map取得
def imnn(): return map(lambda x:int(x)-1, sys.stdin.readline().split()) #整数-1map取得
def fmn(): return map(float, sys.stdin.readline().split()) #浮動小数点map取得
def iln(): return list(map(int, sys.stdin.readline().split())) #整数リスト取得
def iln_s(): return sorted(iln()) # 昇順の整数リスト取得
def iln_r(): return sorted(iln(), reverse=True) # 降順の整数リスト取得
def fln(): return list(map(float, sys.stdin.readline().split())) # 浮動小数点リスト取得
def perm(l, n): return itertools.permutations(l, n) # 順列取得
def perm_count(n, r): return math.factorial(n) // math.factorial(n-r) # 順列の総数
def comb(l, n): return itertools.combinations(l, n) # 組み合わせ取得
def two_distance(a, b, c, d): return ((c-a)**2 + (d-b)**2)**.5 # 2点間の距離
def m_add(a,b): return (a+b) % MOD
def lprint(l): print(*l, sep='\n')

N,A,B = imn()

# 繰り返し2乗法
ans = pow(2,N,MOD)-1

MAX = (10**5)*2

# nCrを求める
def cmb(n,r):
    denominator = numerator = 1
    for i in range(1,r+1):
        denominator = (denominator*i)%MOD
        numerator = (numerator * (n-i+1)) % MOD
    return (numerator * pow(denominator, MOD-2, MOD))%MOD

print((ans-cmb(N,A)-cmb(N,B))%MOD)