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 C (Clang 3.8.0) (judge verdict: Runtime Error)

#include <stdio.h>
#include <math.h>

long long factorial(long long n)
{
    long long a = 1;
    long long i;
    if (n == 0) {
        return 1;
    } else {
    for (i = 1; i <= n; i++) {
        a *= i;
    }
    return a;
    }
}

long long comb(long long a, long long b)
{
    long long c;
    c = factorial(a) / (factorial(b) * factorial(a - b));
    return c;
    }

int main(void){
    long n, a, b, num;
    scanf("%ld", &n);
    scanf("%ld", &a);
    scanf("%ld", &b);
    
    num = (long)pow(2, n) % (long)(pow(10, 9) + 7) - 1 - comb(n, a) - comb(n, b);
    printf("%ld\n", num % ((long)pow(10, 9) + 7));
    
    return 0;
}